ii
• SffiE
TA
PREFACE
THIS text-book on Applied Mechanics is intended for use in the
undergraduate courses in Mechanics in engineering schools. A
knowledge of the principles of General Physics and the Calculus
is assumed. The work in its present form grew out of the author's
attempt to develop the basic principles of the subject in a way
which the average student could easily follow and to present such
illustrations as would show clearly the application of such prin-
ciples to the solution of engineering problems.
Two features may be pointed out in which a departure from the
usual procedure has been made which it is hoped will be advan-
tageous to the student. One of these is the extended use which
has been made of the graphic method of solution. It has been the
author's experience that the graphic method is valuable not only
on account of the ease and rapidity with which it may be applied to
the solution of certain classes of problems, but also on account
of the aid it gives in understanding the algebraic method. The
principles underlying the two methods are developed coordinately
in order to show their relation. The graphic method is used
wherever its application tends to promote clearness.
The other special feature is the large number of illustrative
examples which have been solved in detail to show the relation
between the principle which has been developed and the problems
to which it applies.
More problems are included than can usually be assigned if the
book is to be completed in one semester. Those included in the
articles should always be solved; the general problems at the end of
each chapter may be used as the instructor prefers. The answers
to all problems are given, since it has been found that the average
student will work at a problem with more interest if he has the
answer with which to check his result. Those instructors who
3G5559
vi PREFACE
prefer no answers to be given may make suitable changes in the
data of the problems.
In conclusion the author wishes to thank his colleague, Profes-
sor Richard G. Dukes, for his careful reading of the manuscript
and for his helpful suggestions in regard to form and content.
A. P. POORMAN.
PURDUE UNIVERSITY,
March, 1917.
CONTENTS
PAGE
PREFACE v
GENERAL INTRODUCTION , 1
ART.
1. Definitions 1
2. Fundamental Quantities 1
3. Methods of Analysis of Problems in Mechanics 2
4. Vectors 2
5. Free-body Diagram 3
6. Newton's Three Laws of Motion 3
7. Classification of Forces . . 3
8. Transmissibility of Forces 4
9. Graphical Representation of Forces 4
10. Bow's Notation. . 4
PART I. STATICS
CHAPTER I. CONCURRENT FORCES
11. Definition 5
12. Resultant of Two Forces, Graphically. 5
13. Resultant of Two Forces, Algebraically 6
14. Resolution of a Force into Components 7
15. Resultant of Three or More Forces in a Plane, Graphically 8
16. Resultant of Three or More Forces in a Plane, Algebraically 10
17. Resultant of Three or More Forces in Space, Graphically 13
18. Resolution of a Force into Three Components 14
19. Resultant of Three or More Forces in Space. Algebraically 15
20. Moment of a Force with Respect to a Point 17
21 . Principle of Moments 17
22. Moment of a Force with Respect to a Line 18
General Problems 19
CHAPTER II. PARALLEL FORCES
23. Resultant of Two Parallel Forces. Graphically 23
24. Resultant of Two Parallel Forces, Algebraically 24
25. Resultant of Any Number of Parallel Forces, Graphically 26
26. Resultant of Any Number of Parallel Forces, Algebraically 27
27. Moment of a Couple 29
vii
viii CONTENTS
ART. PAOB
28. Graphic Representation of a Couple 30
29. Composition of Couples 31
30. Resolution of a Force into a Force and a Couple 32
31. Coplanar Parallel Forces in Equilibrium, Algebraic Solution 32
32. Coplanar Parallel Forces in Equilibrium, Graphic Solution 34
33. Equilibrium of Parallel Forces in Space 35
General Problems 36
CHAPTER III. NONCONCURRENT, NONPARALLEL FORCES
34. Composition of Nonconcurrent, Nonparallel Forces in a Plane;
Graphic Methods 39
35. Composition of Nonconcurrent, Nonparallel Forces in a Plane;
Algebraic Method 40
36. Reduction of a System of Forces to a Force and a Couple 41
37. Two-force Members, Three-force Members, etc 42
38. Problems in Equilibrium; Forces in a Plane 44
39. Composition of Nonconcurrent, Nonparallel Forces in Space 49
40. Problems in Equilibrium; Forces in Space 50
41. Cord Loaded Uniformly Horizontally 53
42. The Catenary 56
General Problems. . . 59
CHAPTER IV. CENTROIDS AND CENTER OF GRAVITY
43. Centroid of a System of Forces with Fixed Application Points 65
44. Centroids of Solids, Surfaces and Lines Denned 66
45. Moment with Respect to a Plane 66
4.6. Planes of Symmetry and Axes of Symmetry 67
47. Centroid of a System of Forces with Coplanar Application Points. . 68
48. Centroids of Simple Solids and Surfaces 69
49. Centroids by Integration 70
50. Centroids of Surfaces and Solids of Revolution 73
51. Theorems of Pappus and Guldinus 74
52. Center of Gravity of Composite Body , 75
53. Centroid of Irregular Plane Area 77
54. Center of Gravity by Experiment 77
General Problems . . 78
CHAPTER V. FRICTION
55. Static and Kinetic Friction 82
56. Coefficient of Friction 83
57. Laws of Friction 85
58. Determination of the Coefficient of Friction 85
59. Axle Friction and the Friction Circle 86
60. Least Pull and Cone of Friction 87
CONTENTS ix
ART. PAQB
61. Rolling Resistance 89
62. Friction of Brake on Wheel; Graphic Solution 90
63. Friction on Pivots 92
64. Friction of Belts 94
65. Summary of Principles of Friction 96
General Problems 96
CHAPTER VI. MOMENT OF INERTIA
66. Definition of Moment of Inertia of an Area 99
67. Radius of Gyration 100
68. Sign of Moment of Inertia 102
69. Relation Between Moments of Inertia with Respect to Two Parallel
Axes in the Plane of the Area 102
70. Relation Between Moments of Inertia with Respect to Three Rec-
tangular Axes , . 104
71. Relation Between Polar Moments of Inertia with Respect to
Parallel Axes 104
72. Moment of Inertia of Composite Areas 105
73. Moment of Inertia with Respect to Inclined Axes 106
74. Product of Inertia 107
75. Relation Between Products of Inertia with Respect to Parallel
Axes 108
76. Maximum and Minimum Moments of Inertia 110
77. Moment of Inertia of Mass Ill
78. Relation Between Moments of Inertia of Mass with Respect to
Parallel Axes. . 113
79. Determination of Moment of Inertia by Experiment 115
General Problems . . , 116
PART II. KINETICS
CHAPTER VII. RECTILINEAR MOTION
80. Velocity and Speed 119
81 . Acceleration 120
82. Constant Acceleration 121
83. Falling Bodies, Air Neglected , 122
84. Relation Between Force, Mass and Acceleration 124
85. Effective Forces: D'Alembert's Principle 125
86. Composition and Resolution of Velocities and Accelerations 128
87. Acceleration Varying with Distance: Simple Harmonic Motion. . . . 129
88. Acceleration Varying with Distance: Direct Solution 132
89. Motion in which Acceleration Varies Inversely as the Square of
the Distance 133
90. Relative Motion 134
General Problems. . 137
x CONTENTS
CHAPTER VIII. CURVILINEAR MOTION
ART. PAGE
91. Velocity in Curvilinear Motion 140
92. Acceleration in Curvilinear Motion 140
93. Tangential Acceleration and Normal Acceleration 141
94. Uniform Motion in a Circle 142
95. Simple Circular Pendulum 145
96. Velocity of a Body in a Vertical Curve 147
97. Motion of Projectile, Air Resistance Neglected 149
General Problems 152
CHAPTER IX. ROTATION
98. Angular Displacement 154
99. Angular Velocity 154
100. Angular Acceleration 155
101. Simple Harmonic Motion: Auxiliary Circle Method 156
102. Constant Angular Acceleration 159
103. Variable Angular Acceleration 160
104. Effective Forces on a Rotating Body 160
105. Moment of Tangential Effective Forces 161
106. Resultant of Normal Effective Forces 164
107. Reactions of Supports of Rotating Bodies 166
108. Compound Pendulum 169
109. Center of Percussion 171
110. Centrifugal Tension in Flywheels 171
111. Weighted Conical Pendulum Governor 172
112. Balancing of Rotating Bodies 173
113. Balancing of Bodies in the Same Plane Normal to the Axis of
Rotation 174
114. Balancing of Bodies in Different Normal Planes 175
General Problems 176
CHAPTER X. COMBINED TRANSLATION AND ROTATION
115. Any Plane Motion Equivalent to Combined Translation and
Rotation , 179
116. Resolution of Velocities in Any Plane Motion 180
117. Resolution of Accelerations in Any Plane Motion 181
118. Instantaneous Axis 182
119. Equations of Motion 183
120. Wheel Rolling on Horizontal Plane 185
121. Wheel Rolling on Inclined Plane 186
122. Connecting Rod of Engine. Graphic Solution 187
123. Kinetic Reactions on Connecting Rod 189
124. Kinetic Reactions on Side Rod 191
125. Kinetic Reaction on Unbalanced Wheel 192
126. Balancing Reciprocating Parts 193
CONTENTS xi
ART. PAGE
127. Balancing Both Rotating and Reciprocating Parts 195
128. Balancing of Locomotives 195
General Problems 197
CHAPTER XI. WORK AND ENERGY
129. Work 199
130. Graphical Representation of Work 201
131. Energy 202
132. Relation Between Work and Kinetic Energy 203
133. Kinetic Energy of Translation. Forces Constant 203
134. Kinetic Energy of Translation. Forces Variable 205
135. Kinetic Energy of Rotation 208
136. Kinetic Energy of Rotation and Translation 210
137. Work Lost in Friction 211
138. Braking of Trains 213
139. Power 215
140. Water Power 216
141. Steam Engine Indicator 216
142. Absorption Dynamometer 217
143. Band Brakes 219
General Problems 221
CHAPTER XII. IMPULSE, MOMENTUM AND IMPACT
144. Impulse and Momentum 223
145. Relation Between Impulse and Momentum 224
146. Conservation of Linear Momentum 225
147. Angular Impulse and Angular Momentum 226
148. Conservation of Angular Momentum 228
149. Resultant of Angular Momenta. Gyroscope 230
150. Reaction of a Jet of Water 233
151. Pressure Due to a Jet of Water on a Vane 233
152. Sudden Impulse or Impact 235
General Problems 238
INDEX. . 241
APPLIED MECHANICS.
GENERAL INTRODUCTION.
1. Definitions. Mechanics in general is the science which
treats of the effects of forces upon bodies at rest or in motion.
The complete subject includes many discussions which are of
purely theoretical interest and others which are applicable only
to extremely complex problems. All such discussions are con-
sidered to be beyond the scope of this work, which will include
only those principles of mechanics which are applicable to the
simpler problems of engineering.
Mechanics may be divided into Statics and Kinetics. Statics
treats of bodies at rest or with uniform motion. Kinetics treats
of bodies with variable motion. In this work the kinematic or
pure motion discussion will be included in Kinetics.
2. Fundamental Quantities. Space, time, force and mass are
the fundamental quantities used in mechanics. There are two
systems of units in common use, the centimeter-gram-second (c.g.s.)
system and the foot-pound-second (f.p.s.) system. The foot-
pound-second system is used almost entirely by engineers in
English-speaking countries, so will be the only one used in this
book.
The units of space commonly used in engineering are the foot
and the inch.
The unit of time commonly used is the second.
A force is an action of one body upon another which changes or
tends to change the state of motion of the body acted upon. In
our experience the most common force is the attraction of the
earth for all bodies upon it. The measure of this attraction is
called weight and is determined by means of the spring balance.
Weight varies with the latitude and with the height above sea
level. The unit of force commonly used is the pound.
The mass of a body is the quantity of matter in it, and is con-
stant, regardless of position. Mass may be determined by means
1
2
APPLIED MECHANICS
[INTRO.
of the lever arm balance, or since the acceleration of gravity g
varies the same as the weight W, it may be determined from
simultaneous values of W and g. The unit of mass commonly
used in engineering is one containing g units of weight, or in
symbols,
M = 2.
g
Unless stated otherwise, the value of g will be taken as 32.2 ft.
per second per second. It should be noted that the units of force
and mass in engineering are g times the units used in pure physics.
The reason for this will be explained later.
3. Methods of Analysis of Problems in Mechanics. A
problem in mechanics, as in any other mathematical subject,
consists of a statement of certain known quantities and relations
from which certain other unknown quantities or relations are to be
determined.
There are two methods of analysis of problems, graphic and
algebraic. In the graphic method quantities are represented by
corresponding lines or areas, the relations between them are repre-
sented by the relations of the parts of the figure and the solution
is wholly by determination of the resulting lines or areas. In the
algebraic method quantities are represented by symbols, the
relations between them are shown by signs indicating operations,
and solution is made by algebra and arithmetic. Both methods
will be used in the solution of problems in this book and the student
should soon be able to select the method which is the better suited
to any given problem.
4. Vectors. A vector quantity is a quantity which has direc-
tion as well as amount. Force, velocity, momentum and accelera-
tion are examples. Vector
quantities are represented by
lines called vectors, of definite
length and with definite po-
sition and direction. The
length of the vector repre-
sents to some scale the mag-
nitude of the quantity, its
position shows the line of
FIG. 1
action and an arrowhead shows the direction.
The sum of two or more vectors is found by laying them down
ART. 7] GENERAL INTRODUCTION 3
so as to follow each other in order; then the vector drawn from
the beginning to the end of the system laid down is the sum or
resultant vector. In Fig. 1 the sum of the vectors a} b, c and d is
vector R. The line of action of vector R is not determined in this
figure.
5. Free-body Diagram. The free-body method of analyzing
problems will be used in this work. In applying this method, the
whole body or some part of it is considered as isolated from the
surrounding parts. This "free body" is represented diagrammati-
cally, with the actions upon it of the parts removed indicated by
vectors, known or unknown. From the conditions and forces
known, the unknown relations and forces are determined.
For example, in Fig. 11 (a), the 50-lb. block is to be considered as
the free body. The block is represented diagrammatically by
point A in Fig. ll(b). The attraction of the earth W, the normal
resistance of the plane N, the tangential resistance of the plane
F, and the pull P are represented by their corresponding vectors
and are shown acting upon A. The forces with which the free body
acts upon the parts supposed to be removed are not considered.
6. Newton's Three Laws of Motion. Sir Isaac Newton
formulated the following Laws of Motion, generalized from obser-
vation.
1. Unless acted upon by some force, a body remains at rest or
in uniform motion. This property is called inertia.
2. A body acted upon by a resultant force receives an accelera-
tion in the direction of the force which is proportional to the force
and inversely proportional to the mass of the body.
3. To every action there is an equal and opposite reaction.
7. Classification of Forces. A distributed force is one whose
place of application is an area. A concentrated force is one whose
place of application is so small that it may be considered to be a
point. In many cases a distributed force may be considered as
though it were a concentrated force acting at the center of the area
of contact or at the center of the force system.
Forces are sometimes classified as forces at a distance and forces
by contact. Magnetic, electrical and gravitational forces are
examples of the first. Gravitational force, or the weight of bodies,
is the chief one considered in mechanics. The pressure of steam
in a cylinder and of the wheels of a locomotive on the supporting
rails are examples of forces by contact.
APPLIED MECHANICS
[INTRO.
8. Transmissibility of Forces. By common experience it has
been found that the external effect of a force upon a rigid body
is the same, no matter at what point of the body along the line of
action the force is applied.
9. Graphical Representation of Forces. Since forces are
vector quantities, they are represented graphically by vectors.
For comparatively simple problems only one diagram is used, in
which case each vector shows the line of action, direction and
magnitude of the force represented, as in Fig. 2. For more com-
A
B
Space and Force Diagrams
Combined
Fia. 2
9
Space Diagram
FIG. 3
G,
E
F
Force Diagram
plicated problems two diagrams are used, the space diagram
showing the lines of action of the forces, and the force diagram
showing the magnitude of the forces, as in Fig. 3. The direction
of the forces may be shown in either diagram but is usually shown
in both.
10. Bow's Notation. In the graphical work Bow's Notation
will be used in all except very simple problems. In the space
diagram each space from the line of action of one force to that of
the next one is lettered with a lower case letter. The line of
action be in Fig. 3 is the line between space b and space c. The
corresponding upper case letters are placed at the ends of the
corresponding vector in the force diagram. Thus BC represents
in amount the force acting along line be in the space diagram.
PART I. STATICS.
CHAPTER I.
CONCURRENT FORCES.
11. Definition. A concurrent system of forces is one in which
the lines of action of all the forces meet in a common point. If all
of the forces in the system are in the same plane, it is called also a
coplanar system.
12. Resultant of Two Forces, Graphically. The Parallelogram
Law. If two concurrent forces are represented by their vectors, both
of which are directed away from their point of intersection, the diagonal
of the completed parallelogram drawn through their point of inter-
section represents their resultant.
In Fig. 4, let vectors MN and
KL represent two forces whose
lines of action intersect at 0. ^
By the principle of Art. 8 the .
forces may be transmitted along
their lines of action until they
are in the positions OA and OB.
Line AC is drawn parallel to OB
and line BC parallel to OA, to complete the parallelogram OACB.
The diagonal OC is the vector sum of the two vectors OA and OB
and represents the resultant of the two forces.
If the two vectors had been placed so that they were both
directed toward their point of intersection, their resultant vector
would have been the same.
The Triangle Law. If two concurrent forces are represented by
their vectors laid down in order as
the two sides of a triangle, the third
side of the triangle drawn from the
initial point to the final point
FIG. 5 represents their resultant. Fig. 5
shows the two possible solutions for obtaining the resultant of the
5
6
APPLIED MECHANICS
[CHAP, i
forces KL and MN of Fig. 4. In the lower triangle vectors OA
and AC represent the two forces, and vector OC represents their
resultant. In the upper triangle vectors OB and BC represent the
two forces and vector OC represents their resultant as before.
These diagrams are force diagrams only. The line of action of
the resultant must pass through the point of intersection of the
two lines of action in the space diagram.
If two forces have the same line of action, the resultant is the
algebraic sum of the two. If they are equal and opposite, their
sum is zero and the two forces are said to
be in equilibrium.
Conversely, if two forces are known to be
in equilibrium, they must be equal, opposite
and colinear.
J
4lbs.
n
1
f8
Ibs.
V,
6/bs.
FIG. 6
Problem 1. Fig. 6 represents a body three feet
square to which forces are applied as shown.
Combine the 4-lb. force with each of the others
in turn by means of the parallelogram law.
Ans. With 84b. force, R = 11.86 Ibs.; 0 (with X) = 77° 40'.
With 6-lb. force, R = 9.26 Ibs.; 0 = 62° 48'.
With 10-lb. force, R = 10.77 Ibs.; 0 = 158° 12'.
With 3-lb. force, R = 1 lb.; 0 = 90°.
Problem 2. Combine the 3-lb. force of Fig. 6 with each of the others in
turn by means of the triangle law.
Ans. With 8-lb. force, R = 5.25 Ibs.; 0 = 61° 02'.
With 6-lb. force, R = 4.42 Ibs.; 0 = 16° 20'.
With 10-lb. force, R = 10.44 Ibs.; 0 = 196° 40'.
With 4-lb. force, R = 1 lb.; 0 = 90°.
13. Resultant of Two Forces, Algebraically.
nometry of Fig. 7,
From the trigo-
and
FIG. 7
OR2 = OP2 -f OQ + 2 OP X PQ cos «,
,~AR OQ sin a
6 = tan-1 = = tan-1 — —
OA OP + OQ cos a
In the special cases when a = 0°, 90° or 180°, these expressions
are much simplified.
ART. 14]
CONCURRENT FORCES
For a = 0°, OR = OP + OQ, and 0 = 0°.
For a = 90°, OR =
P* + OQ\ and 6 = tan"1
For a = 180°, OR = OP-OQ, and 6 = 0° or 180°.
Problem 1. Find the resultant of the 6-lb. force and the 4-lb. force of
Fig. 6 by the method of this article.
Problem 2. Find the resultant of the 6-lb. force and the 10-lb. force of
Fig. 6. Ans. R = 7.15 Ibs.; 0 with 6-lb. force = 98° 36'.
Problem 3. Find the resultant of the 4-lb. force and the 10-lb. force of
Fig. 6. Ans. R = 10.77 Ibs.; d = 158° 12'.
14. Resolution of a Force into Components. By reversing the
parallelogram law or the triangle law, any force may be resolved
into two components. Let the vector AB in Fig. 8 represent the
force. Through any point C draw the two lines AC and CB.
The vectors AC and CB represent the components of AB. C may
be any point, so any number of pairs of components may be
obtained. The components usually desired are those parallel to
certain given axes, such as ACi and CiB, horizontal and vertical
respectively.
FIG. 8
FIG. 9
In Fig. 9, let F be the force, Fx and Fy its components parallel to
the X and Y axes respectively, and a the angle the force makes
with the X axis.
By trigonometry,
Fx = F cos a and Fy = F sin a.
The components Fx and Fy are called the projections of the force
F upon the X and Y axes.
A force may be resolved into its components at any point along
its line of action.
Problem 1. In Fig. 9, let F = 100 Ibs. and a = 30°. Determine Fx
and Fy. Ans. Fx = 86.6 Ibs. Fy = 50 Ibs.
8
APPLIED MECHANICS
[CHAP, i
(b)
FIG. 10
Problem 2. Let F = 40 Ibs. and a = 45°. Determine Fx and Fy.
Ans. Fx = 28.28 Ibs. Fv = 28.28 Ibs.
Problem 3. Let F = 120 Ibs. and a = 110°. Determine Fx and Fy.
Ans. Fx= - 41.04 Ibs. Fy = 112.8 Ibs.
Problem 4. Let F = 500 Ibs. and a = 30°. Resolve F into two com-
ponents, one horizontal, the other at an angle of 120° with the positive end of
the X axis. Ans. 577.3 Ibs. horizontal. 288.7 Ibs. at 120°.
15. Resultant of Three or More Forces in a Plane, Graphi-
cally. By an extension of the Triangle Law of Art. 12, the
c resultant of any number of con-
current forces may be found.
In Fig. 10, AB and BC are com-
bined into their resultant AC;
AC and CD are combined into
their resultant AD; and finally
AD and DE are combined into
their resultant AE, which is
therefore the resultant of the
entire system. Its line of ac-
tion passes through 0, the common point of the system.
It will be noticed that in making the solution, vectors AB, BC,
CD and DE may be laid down in order, then the final resultant AE
may be drawn without using the intermediate resultants AC and
AD. Fig. 10(b) is called the Force Polygon. If the forces are
taken in any other order, a different force polygon will be obtained,
but the same resultant.
If the last force closes at the starting point, the resultant R = 0
and the system is in equilibrium. In any case, another force equal
and opposite to R through the common point of the system will
hold it hi equilibrium.
Conversely, // the force system is in equilibrium, the force polygon
must close.
In the special case of three forces in equilibrium, the following
important principle also applies: If a force system of three non-
parallel forces is in equilibrium, they must meet in a common point.
For if any two of the forces are combined into their resultant, this
resultant acts through their point of intersection. Then in order
for the third force to balance this resultant and hence the other
two forces, it must also pass through their point of intersection
and be equal and opposite to their resultant.
ART. 15]
CONCURRENT FORCES
9
These principles are used in the solution of problems in which
the body and hence the force system acting upon it is known to be
in equilibrium, but some of the forces are unknown.
EXAMPLE 1.
A block weighing 50 Ibs. is lying on a plane inclined at an angle of 30°
with the horizontal. If the friction of motion is 15 Ibs., what force P par-
allel to the plane will be required to move the block uniformly up the plane?
Solution: — Fig. 11 (a) shows the block resting upon the 30° plane with
the force P acting upon it. In the free-body diagram, Fig. ll(b), the actions
of all the surrounding parts upon the block as the free body are shown
diagrammatically. The weight is a force of 50 Ibs. vertically downward.
The total reaction of the plane is resolved into two components, one, F,
parallel to the plane (friction), the other, N, normal to the plane. The
friction is a force of 15 Ibs. downward along the plane. (Friction of the
adjoining surfaces always opposes the motion of the free body.) N, the
normal resistance of the plane, is unknown in amount and P is the unknown
force asked for. Both N and P, however, are known in direction.
Since the body is in uniform motion, the forces are in equilibrium, so
graphically the force polygon must close. This is drawn by laying down to
scale vector W of the 50-lb. force and vector F of the 15-lb. force of friction,
Fig. ll(c). Two other vectors, one parallel to N, Fig. ll(b), and one parallel
to P, must close at the initial point. These are drawn and their lengths
scaled. N = 43.3 Ibs.; P = 40 Ibs.
EXAMPLE 2.
A boom 30 ft. long, Fig. 12 (a), weighs 1200 Ibs. and is supported in a
horizontal position by a cable BC at an angle of 30° with the horizontal.
Determine the tension in the cable BC and the amount and direction of the
hinge reaction at A.
Solution: — It is known that the boom is in equilibrium under the action of
three forces, its weight vertically downward through the center of gravity D,
the tension in the cable BC, and the hinge reaction through A, The line of
action of the weight intersects BC at E. By the principle of Art. 15 the line
of action of the hinge reaction at A must also pass through E, so its direction
is determined, and is at an angle of 30° with the horizontal. The free-body
10
APPLIED MECHANICS
[CHAP, i
diagram for the boom is shown in Fig. 12(b) and the solution of the force
triangle in Fig. 12(c). The tension in the cable and the hinge reaction at
A scale 1200 Ibs. each.
T
(C)
It will be noticed that the force triangle formed of the vectors P, T and W
is similar to the triangle AEB on the truss, since their sides are mutually
parallel. As an alternative to drawing the force diagram accurately and
scaling the values, the two unknown forces may be calculated from the pro-
portionality of sides of the two similar triangles, since the dimensions of the
triangle AEB are known.
P AE
W~ AB*
17 09
p-wooxfif-im
Problem 1. In Fig. 11 (a), let P act horizontally and let the friction F be
20 lbs. Solve for the force P necessary to move the block uniformly up the
plane. Ans. P = 52 lbs.
Problem 2. In Problem 1 solve for the horizontal force P necessary to
allow the block to move uniformly
down the plane.
Ans. P = 5.77 lbs.
Problem 3. A weight of 150 lbs.
suspended by a cord is pulled to one
side by a horizontal force of 40 lbs.
What is the tension in the supporting
cord and its angle with the vertical?
Ans. 155.2 lbs.; 6 = 14° 56'.
Problem 4. Determine the tension
in each cord and the value of the angle 9
in the system of cords shown in Fig. 13.
\ = 136.7 lbs. T2 = 70.7 lbs. Tz = 36.6 lbs.
T, = 100 lbs. T6 = 122.5 lbs. 6 = 15°.
16. Resultant of Three or More Forces in a Plane, Algebrai-
cally. As in the preceding article, the principles of Art. 13 may
be extended to the case of three or more forces. Any two may be
combined, then the resultant of these with a third and so on.
This method is cumbersome and will be found of little use.
ART. 16]
CONCURRENT FORCES
11
The principle of resolution and recomposition is more easily
applied. By the principles of Art. 14 each force may be resolved
at the common point of intersection into two forces along the two
rectangular axes. All the X components may next be combined
into one force called 2FX and all the F components into one force
called SFy. These two forces may then be combined into the
final resultant of the system R.
R =
A concurrent system of forces in a plane is in equilibrium if
R = 0, which can be only if ZFX = 0 and ^Fy = 0.
Conversely: // a system of concurrent forces is in equilibrium,
2FX = 0 and 2Fy = 0.
This principle is used in the solution of problems in which a
force system is known to be in equilibrium, but some of the forces
are unknown.
EXAMPLE 1.
Determine the amount and direction of the resultant of the four forces
represented in Fig. 14.
Solution: — Each force in order is replaced by its X and Y components,
Fx = F cos a. and Fv = F sin «, as tabulated below.
F (Ibs.) Fx = F cos a
F, = 10 10 X 0.866 = 8.66
F2 = 80 80 X 0 = 0.
F8 = 50 50 X(-0.9848) = -49.24
F, = 100 100 X 0.707 = 70.7
?FX = +30.12
Fy = F sin a
10 X 0.5 = 5.
80 X 1 =80.
50 X (-0.1736) 8.68
100 X (-0.707) = -70.7
VF = + 5.62
4-
80/bs.
./f
0
./^ o
X_ ^ X
F5 =50 Ibs.
\ f5°
FIG. 14
FIG.
The X components are added algebraically and give +30.12 Ibs. for the
resultant force along the X axis, SFX. In the same way the Y components are
added algebraically and give +5.62 Ibs. for the resultant force along the Y
axis, SFy. The final resultant is obtained by taking the square root of the sum
of the squares of the two rectangular components.
12
APPLIED MECHANICS
[CHAP, i
R = V30.122 -f 5.622 = V938.8 = 30.64 Ibs., as shown in Fig. 15.
angle 6 with the X axis is given by the expression
5.62
The
e = tan-1
tan"1
30.12
10° 35'.
EXAMPLE 2.
A cast iron sphere 1 ft. in diameter rests in an 8 in. X 8 in. angle, one leg
of which is at an angle of 30° with the horizontal, as shown in Fig. 16 (a).
What are the pressures at A and B1
44 1
Solution: — The volume of a sphere = ^ ?rr3 = - X IT X ~ = 0.5236 cu. ft.
66 o
Weight = 0.5236 X 450 = 235.6
Ibs. Fig. 16(b) is the free-body
diagram for the sphere. The
weight is a force of 235.6 Ibs.
vertically downward, represented
by vector W. Since there is no
tendency for the sphere to move
at the points of contact, the pres-
sure of each side of the angle
upon the sphere is normal to the
(a)
FIG. 16
(b)
surface at that point. These pressures are represented by vectors PA and
PB.
If the summation of horizontal forces in the free body diagram is equated
to zero and the summation of vertical forces is equated to zero, each equation
will contain two unknown quantities and the two must be solved simultane-
ously in order to determine the unknown quantities.
If the summation of forces parallel to PA is equated to zero, the equation
will contain only one unknown quantity which is immediately determined.
PA -TFsinSO0 =0.
PA = 117.8 Ibs.
Similarly, if the summation of forces parallel to PB is equated to zero, PB
is determined.
Pfi-TFcos30° =0.
PB = 204 Ibs.
Problem 1. Check the results obtained for Example 1 above, using the
line of action of the 100-lb. force as the X axis.
Problem 2. The system of forces shown in Fig. 17 is known to be in equi-
librium. Determine the amounts of the unknown forces 7\ and Tz.
Am. Ti = 3010 Ibs. T2 = 985 Ibs.
1,400 Ibs.
\50Qlbs.
,3,600/bs.
FIG. 17
FIG. 18
ART. 17] CONCURRENT FORCES 13
Problem 3. A wheel 2 ft. in diameter carries a load of 1000 Ibs. as shown
in Fig. 18. What horizontal force P applied at the axle is necessary to start
it over an obstacle 6 inches high? What force is necessary if the wheel is 4 ft.
in diameter?
Ans. P = 1732 Ibs. P = 882 Ibs.
Problem 4. A boom P, Fig. 19, is held by a
pin A at the wall CA, and a cable T fastened at
the outer end and running up to the wall at an
angle of 30° with the boom. A load of 1000 Ibs.
is carried at point B. Determine the compres- K— —10— —
sion in P and the tension in T. 1,000 Ibs.
Ans. P = 1732 Ibs. T = 2000 Ibs. FIG. 19
Problem 5. Solve Example 2, Art. 15, by the method of this article.
17. Resultant of Three or More Forces in Space, Graphically.
Any number of concurrent forces in space may be combined
graphically by a slight extension of the Triangle Law, Art. 12.
Any two of the forces may be combined into their resultant, then
this resultant may be combined with the third force, which in
general will not be in the same plane with the first two, and so on
until the final resultant is obtained. When the system consists
of three forces mutually at right angles, the resultant is the diag-
onal of the rectangular parallelepiped constructed upon the three
vectors. If the force polygon (in space) closes, the resultant is
zero and the forces are in equilibrium.
Conversely: // any system of concurrent forces in space is in
equilibrium, the force polygon closes.
Also, // the force polygon in space closes, the projection of the
force polygon on each of the three reference planes closes.
In the solution of problems in equilibrium, the projection of the
given system upon some reference plane is drawn and from the
fact that the projection of the force polygon must close, the un-
known forces are determined.
EXAMPLE.
A shear-legs crane has dimensions and load as shown in Fig. 20 (a). Deter-
mine the stress in AE and the stress in AB.
Solution: — Take as the plane of projection the vertical plane AEF. The
force system projected upon this plane is shown in Fig. 20(b), which may also
be called the free-body diagram of point A. In this projection the forces
BA and DA are superimposed. In Fig. 20 (c) the graphical solution of this
projected system" is made. The vector T gives the stress in AE and scales
14
APPLIED MECHANICS
[CHAP, i
15,120 Ibs. The vector P' gives the sum of the projected values of the stresses
in BA and DA, 31,500 Ibs.
(a)
(c)
FIG. 20
In order to determine the stresses in B A and DA, a view in the plane
ABD is taken, Fig. 21 (a). Vector P' acts along CA and is really the resultant
of the stresses in BA and DA. In order to determine these
stresses, vector P' is laid down parallel to CA, as in Fig.
21(b). Through its initial point a line is drawn parallel to
AB and through its final point a line is drawn parallel to
AD. Their intersection determines the length of the vectors
P, P, which represent the stresses in B A and DA. The
scaled value of each is 15,400 Ibs.
Problem 1. A weight of 50 Ibs. is hung from a hori-
zontal ring 4 ft. in diameter by means of three cords each
4 ft. long. On the ring the cords are placed 120° apart.
Find the tension in each cord. Ans. 19.22 Ibs.
Problem 2. Solve Problem 1 if two of the cords are 90° apart and the point
of attachment of the third bisects the remaining arc.
Ans. 16.9 Ibs. 16.9 Ibs. 23.9 Ibs.
18. Resolution of a Force into Three Components. The most
common case of the resolution of a force into three components is
that in which the components are parallel respectively to the three
rectangular axes.
y
FIG. 22
FIG. 23
If the angles between the force and the axes are given, a with X,
ft with Y and y with Z, the algebraic method of resolution is more
easily applied than the graphic method.
Fx = F cos a, Fy = F cos ft and FM = F cos y
ART. 191
CONCURRENT FORCES
15
are the rectangular components along the three axes, as shown in
Fig. 22.
If the rectangular coordinates of two points on the line of action
of the force are given instead of the angles a, |8 and 7, their cosines
may be computed from the dimensions given and the resolution
made algebraically. The graphic method, however, is very readily
employed in this case.
Let the force be F, acting diagonally from A to C, Fig. 23.
Pass a vertical plane ABCD through the force. Fig. 24(a) shows
F1
Co)
G B
H
FIG. 24
the view normal to this plane, in which vector F is resolved into
vector F' horizontal and Fy vertical. Fy is one of the components
desired. In the horizontal plane AEGH, the top view of which is
shown in Fig. 24 (b), the force F' is resolved into its two components
parallel to the other axes, Fx and Fz.
Problem 1. In Fig. 23, let GE = 6 ft., GB = 3 ft., GH = 4 ft., GO = 4 ft.,
and F = 20 Ibs. Resolve force F into its three rectangular components, Fx,
Fya,ndFz. Ans. -9.38 Ibs. -12.5 Ibs. -12.5 Ibs.
Problem 2. If a force of 100 Ibs. has line of action OA, Fig. 23, determine
its three rectangular components, Fx, Fy and Fz.
Ans. 72.8 Ibs. 48.5 Ibs. 48.5 Ibs.
Problem 3. In Fig. 22, F = 1000 Ibs., a. = 45°, 0 = 64° 50' and 7 = 55° 30'.
Determine its three rectangular components, Fx, Fy and Fz.
Ans. 707 Ibs. 425.3 Ibs. 566.4 Ibs.
19. Resultant of Three or More Forces in Space, Algebrai-
cally. If F1} F2, etc., Fig. 25, are the forces, at angles (on, fa, 71)
(«2, ft, 72), etc., with the X, Y and Z axes respectively, the
resultant may be determined as follows:
At the common point of intersection each force may be resolved
into its X, Y and Z components. The X components may
be added algebraically into HFX, the Y components into
16
APPLIED MECHANICS
[CHAP, i
and the Z components into SF2. The final resultant of these
three is,
R =
FIG. 25
The angles that R makes with the three axes are given by the
following expressions:
a = cos
= cos-'
7 =
D ) K 7? ' ' E>
ifc X£ • <f*
If # = 0, SFX = 0, SFtf = 0 and SF2 = 0, and the force system
is in equilibrium.
Conversely, // a system of concurrent forces is in equilibrium, the
summation of forces along any line equals zero.
Also, The projection of any system of concurrent forces in equilib-
rium upon any plane constitutes a system in equilibrium.
EXAMPLE.
Fig. 2Q represents a hay stacking outfit. With a load of 1000 Ibs. at the
middle and a sag of 4 ft. below the horizontal, what are the stresses Ti, T2
andP?
y
FIG. 26
Solution: — Consider first the cable ACB and the load as the free body.
The stresses T2 and the weight of the load constitute a coplanar system of
ART. 21] CONCURRENT FORCES 17
forces in equilibrium. With a sag of 4 ft. at C, the length AC = 25.32 ft.
Equation *2FV — 0 gives
1000.
T2 = 3165 Ibs.
Length AO = 30 sin 60° = 25.98 ft.
Length OD = V502 - 25.9S2 = 42.7 ft.
The four forces at A constitute a concurrent system in equilibrium. Equa-
tion SFz = 0 gives
3165 * 2-1-2 -frr- = °'
7\ = 3660 Ibs.
Equation ZFy = 0 gives
OK QQ A
3660 X ~ + 3165 X ~ 2 P X 0.866 = 0.
P = 1388 Ibs.
Problem 1. A tripod with legs 8 ft. long is set up on a level floor with each
leg at the vertex of an equilateral triangle whose sides are each 4 ft. long.
What are the stresses in the legs caused by a load of 100 Ibs. on top?
Arcs. 34.8 Ibs. each.
Problem 2. Solve the Example in Art. 17 by the method of this article.
20. Moment of a Force with Respect to a Point. The moment
of a force with respect to a point is the product of the force and the
perpendicular distance from its line of action to the point. This
perpendicular distance is called the arm of the force; the point is
called the center of moments.
Let F, Fig. 27, be the force, d its arm and M0 the moment of the
force with respect to point 0. Then
Mo = Fd.
The moment of a force is the measure of its
tendency to rotate the body upon which it acts
around an axis through the center of moments,
normal to the plane through the force and its arm. FlQ> 27
Unit of Moment. Moment is measured in terms of the units of
length and force used; as, foot-lbs., inch-lbs., inch-tons, etc.
Sign of Moment. Moments tending to produce rotation counter-
clockwise are commonly called positive, those clockwise, negative.
The opposite notation may be used if kept consistently throughout
the problem.
21. Principle of Moments. The algebraic sum of the moments
of two concurrent forces with respect to a point in their plane is equal
to the moment of their resultant with respect to the same point.
18 APPLIED MECHANICS [CHAP, i
Let P and Q, Fig. 28, be the forces concurrent at A, R their
resultant and 0 any point in their plane. Draw AO and produce
it to F. From the ends of P and
R drop perpendiculars CE, DF
and CG. Also drop perpendiculars
p, q and r from 0 to the forces P,
Q and R respectively. Let a, |3
£ ?~~ and 6 be the angles between the
FIG. 28 line AO and the forces P, Q and
R respectively. Then
FD = FG +
so R sin 0 = P sin a + Q sin 0.
This equation multiplied by OA becomes
R - O4 sin 0 = P . OA sin a + Q • OA sin ft
or
Rr = Pp + Qq.
The student should supply the proof when 0 is between P and
R or between Q and #.
From the principle above, the moment of a force with respect to
a point is equal to the sum of the moments of its X and Y com-
ponents with respect to the same point. It is often simpler to
compute this sum than the moment of the force itself.
This proof may be extended to the case of three or more con-
current forces. The statement of the general case, then, is as
follows: The moment of the resultant of any number of concurrent
forces in a plane with respect to any point in that plane is equal to the
algebraic sum of the moments of the forces with respect to the same
point.
The above demonstration is commonly called "Varignon's
Theorem."
Problem 1. In Fig. 28, let P = 100 Ibs.; Q = 60 Ibs.; angle between P
and Q = 60°; angle BAO = 80°; OA = 4 ft. Determine the resultant R
and the moment of this resultant with respect to point 0.
Am. R = 140 Ibs. M0 = 373.2 ft.-lbs.
22. Moment of a Force with Respect to a Line. The moment
of a force with respect to a line parallel to it is zero, since there is no
tendency for the force to rotate the body upon which it acts about
that line as an axis. The moment of a force with respect to an
ART. 221
GENERAL PROBLEMS
19
axis intersecting it is zero, since the moment arm is zero. The
moment of a force with respect to an axis in a plane perpendicular
to the force is equal to the product of the force and the perpendicu-
lar distance from the force to the axis.
If the axis is not in a plane perpendicular to the force, the force
may be resolved at any point into two rectangular components,
one parallel to the axis, the other perpendicular to a plane con-
taining the axis. The moment of the original force with respect to
the axis is equal to the moment of the perpendicular component
alone, since the moment of the component parallel to the axis is
zero.
Another method is to resolve the force into three mutually
rectangular components, one of which is par-
allel to the axis and hence has no moment
with respect to it. The sum of the moments
of the other two components gives the mo-
ment of the original force.
Problem 1. In the force system shown in Fig. 29,
determine the resultant moments with respect to the
X, Y and Z axes. Each side of the parallelepiped is
3 ft. long.
Am. Mz = - 2.1 ft.-lbs. My = - 141.4 ft.-lbs. Mz
80/bs.
FIG. 29
- 238.6 ft.-lbs.
GENERAL PROBLEMS.
Note: — In many cases in this and the following lists of general problems
the student has a choice of methods of solution, so care should be taken to
choose the method best adapted to the problem. If there is no figure to illus-
trate the problem, a sketch is a great help in the solution.
The free-body diagram should always be drawn.
Problem 1. A weight of 50 Ibs is supported by two cords, one at an angle
of 30° with the horizontal, the other at an angle of 45° with the horizontal.
I hid the tension in the cords. Ans. 36.6 Ibs. 44.8 Ibs.
Problem 2. A slack wire performer weighing 150 Ibs. stands in the middle
of a wire 30 ft. long and depresses it 6 ft. What is the tension in the wire due
to the man's weight? Ans. 187.5 Ibs.
20
APPLIED MECHANICS
[CHAP, i
Problem 3. A picture weighing 30 Ibs. is hung by an endless wire passing
over a hook at A and through screw rings at four points as shown in Fig. 30.
What is the tension in the cord and the resultant
pull on each screw?
Ans. Tension = 16.8 Ibs. Top screws, 7.71 Ibs. at
13° 20' with horizontal. Bottom, 23.75 Ibs. at 45°
with horizontal.
Problem 4. Boatmen say that a mule can pull a
heavier tow boat with a long hitch than with a short
one, while teamsters claim that in their work the
opposite is the case. Explain.
Problem 5. A force of 500 Ibs. acts at an angle
of 30° with the horizontal. Find its vertical and
FIG. 30
horizontal components.
Solve also if the force acts at an angle of 15° with the horizontal.
Ans. 250 Ibs. 433 Ibs. 129.4 Ibs. 483 Ibs.
Problem 6. A horizontal force of 180 Ibs. is pulling a body up a 20° plane.
Find the components parallel to the plane and perpendicular to the plane.
Ans. 169.2 Ibs. 61.6 Ibs.
Problem 7. A guy wire to a smokestack makes an angle of 40° with the
ground. When the tension in it is 10,000 Ibs., what are its vertical and hori-
zontal components?
The other two wires are at 120° with the first, but are at 45° and 50° with
the ground. What is the tension in each and what vertical compression is
caused in the stack by the three?
Ans. 6428 Ibs. 7660 Ibs. 10,830 Ibs. 12,910 Ibs. 23,228 Ibs.
Problem 8. The upward reaction of the pedestal of a bridge upon the end
pin is 8000 Ibs., as shown in Fig. 31. Consider the actions of the end post and
lower cord to be axial and determine the stress in each.
Ans. 9240 Ibs. comp. in post. 4620 Ibs. tens, in lower chord.
FIG. 31
B C
FIG. 32
Problem 9. Six cylinders of equal size, each weighing 1000 Ibs., are piled
up as shown in Fig. 32. Find the pressures at A, B and C.
Ans. 577 Ibs. 2000 Ibs. 2000 Ibs.
Problem 10. Three uniform spheres, each weighing 12 Ibs., just fit into a
triangular box with vertical sides. A fourth sphere of the same size and weight
is placed on top of the three. What pressure does each exert on the box at the
three points of contact? Ans. 16 Ibs. on bottom. 2.83 Ibs. at each side.
GENERAL PROBLEMS
21
Problem 11. A wheel 3 ft. in diameter carries a load of 500 Ibs. What is
the amount and direction of the least force P which will start it over an ob-
struction 6 inches high? Am. 373 Ibs. at 48° 10' with the horizontal.
(Note: — Solve graphically. The weight is represented by a vector, known
completely. The direction of the reaction of the obstruction is radial. The
force P called for is given by the least vector which will close the triangle.)
Problem 12. One end of a horizontal rod 6 ft. long is pinned to a vertical
wall and the other is supported by a cord passing up to the wall at a point 8 ft.
above the pin. A load of 400 Ibs. is hung at the
outer end. What is the tension in the cord and
the compression in the rod?
Ans. Tension = 500 Ibs. Compression = 300 Ibs.
Problem 13. If the weight is hung at the mid-
dle of the rod of Problem 12, what is the tension
in the cord? What is the amount and direction of
the pin pressure at the wall?
Ans. Tension = 250 Ibs. R = 250 Ibs.
e with hor. = 53° 08'.
Problem 14. Fig. 33 shows a simple derrick.
What are the stresses in AB and AC1
Ans. AB = 2500 Ibs. tens. AC = 2500 Ibs. comp.
Problem 16. In the crane shown in Fig. 34, determine the compression in
BD and the amount and direction of the pin pressure at A.
Ans. 1300 Ibs. 900 Ibs. 43° 50' with horizontal.
2,000lbs.
FIG. 33
FIG. 34
FIG. 35
Problem 16. The boom of the stiff-leg derrick shown in Fig. 35 has a range
of position in a vertical plane from the horizontal to within 20° of the vertical.
Determine the position of the boom for the maximum stress in BC. For this
position of the boom in the vertical plane, determine the value of angle a for
the maximum compression in BE. Same for the maximum tension. If the
load at C is 2400 Ibs., determine the maximum stresses in BC, AC, BE
and BA.
Ans. BC = 4000 Ibs. AC = 3200 Ibs. BE = 4525 Ibs. BA = 6920 Ibs.
22
APPLIED MECHANICS
[CHAP, i
Problem 17. In the Example of Art. 17 resolve each force as there ob-
tained into its X, Y and Z components.
Ans. For AE, Fx = -12,080 Ibs. Fy = -9120 Ibs. Fz = 0.
For AB, Fx = 5790 Ibs. Fy = 13,970 Ibs. Fz = 3080 Ibs.
Problem 18. A shear legs crane is loaded as shown in Fig. 36. Find the
stresses in AC and DC.
Ans. DC = 853 Ibs. tens. AC = 825 Ibs. comp.
Problem 19. The pressure of the steam on the
piston of an engine is 15,000 Ibs. Neglecting friction,
what is the pressure on the guide and the compres-
sion in the connecting rod when the connecting rod
makes an angle of 15° with the direction of the pis-
ton rod? Ans. 4020 Ibs. 15,520 Ibs.
Problem 20. What weight can be drawn up a
smooth plane with a slope of 1 in 10, by a force of 50 Ibs. acting parallel to the
plane? How much by 50 Ibs. acting horizontally?
Ans. 502.5 Ibs. 500 Ibs.
Problem 21. A tripod A BCD, Fig. 37, with vertex at A has leg AC 27.5 ft.
long, leg AB 25 ft. long and leg AD 30 ft. long. It is placed on level ground
with distance BC 20 ft., distance CD 30 ft. and distance BD 25 ft. With 6000
Ibs. vertical load at A, what is the compressive stress in each leg? (Solve
graphically.) Ans. AB = 2900 Ibs. AC = 2100 Ibs. AD = 1950 Ibs.
FIG. 36
FIG. 37
FIG. 38
Problem 22. Determine the resultant of the five forces shown in Fig. 38,
both in amount and direction.
Ans. R = 74.1 Ibs. « = 117° 52'. 0 = 135° 30'. 7 = 121° 30'.
Problem 23. In the force system shown in Fig. 38, determine the moment
of each force with respect to each axis. Each side of the cube is 5 ft. long.
Ans. F Mx My
(ft.-lbs.)
+42.90
F
(Ibs.)
10
20
30
40
50
MX
(ft.-lbs.)
-25.75
+ 100
-23.40
+ 196
-176.75
-100
+ 117
-39.25
0
Mz
(ft.-lbs.)
-17.15
0
-93.85
-156.80
+ 176.75
CHAPTER II.
PARALLEL FORCES.
23. Resultant of Two Parallel Forces, Graphically. The
graphic method of finding the resultant of forces, as used in Art. 12,
must be modified in the case of parallel forces, since they do not
intersect. By one method, one of the forces is resolved into two
components, then the resulting system of three forces is combined
into their resultant.
There are two cases, one in which the two forces are in the same
direction, as in Fig. 39, the other in which they are in opposite
a b
(a} Space Diagram
(b) Force Diagram
FIG. 39
directions, as in Fig. 40. The solution is simplified by using both
the space and force diagrams. In either figure let AB and BC
(a) Space Diagram
(b) Force Diagram
FIG. 40
represent the forces, acting along ab and be. At any point n,
resolve AB into any two convenient components, AO and OB,
acting along ao and ob. Where ob intersects be, at m, combine
23
24
APPLIED MECHANICS
[CHAP, ii
OB and BC into their resultant OC acting along oc. The system
now consists of forces AO and OC, concurrent at p in the space
diagram. Combine these into their resultant AC which acts
through the point of intersection p parallel to the original forces.
The amount and the direction of the resultant are given by the
algebraic sum of the original forces.
The original polygon in the force diagram is called the force
polygon, point 0 is called the pole and the auxiliary lines OA, OB,
etc., are called rays. The polygon in the space diagram is called
the funicular polygon, and the component parts of it, oa, ob, etc.,
are called strings.
It will be noticed that in the case in which two parallel forces
are oppositely directed, the resultant is not between the two, but
is outside on the side of the larger force.
If the two forces are opposite in direction and equal in amount,
the resultant is zero and point C, Fig. 40, falls at A in the force
diagram. Then line oc will be parallel to oa, hence the forces OC
and OA cannot be combined into a single force. Such a system is
called a couple and will be discussed in Arts. 27-30.
Problem 1. In Fig. 39, find the amount and position of the resultant if
force AB is 8000 Ibs., force BC is 2000 Ibs. and the distance between their lines
of action is 5 ft.
Ans. R = 10,000 Ibs., 1 ft. from db.
Problem 2. Determine the amount and
position of the resultant of the two wheel
loads shown in Fig. 41.
Ans. R = 2600 Ibs., 3.85 ft. from larger
wheel.
Problem 3. In Fig. 40, find the amount
IJbOOIbs.
1.000 Ibs.
t
10' ->)
FIG. 41
and position of the resultant if force AB is 500 Ibs., force BC is 300 Ibs. and
the distance between their lines of action is 3 inches.
Ans. R = 200 Ibs. downward, 4.5 in. to the left of db.
Problem 4. The horizontal pressures of the support
upon the wheels of an elevator car are as shown in Fig.
42. Find the amount, direction and position of the
resultant.
Ans. R = 150 Ibs. to the right, 2.67 ft. below 250-lb.
force.
24. Resultant of Two Parallel Forces, Al- ®°
gebraically. As stated in Art. 23, the result- F10- 42
ant of two parallel forces is given in amount and direction by
the algebraic sum of the component forces. It remains, then,
ART. 24]
PARALLEL FORCES
25
to find the position of the line of action. Let AB and BC, Fig.
43, be the two forces and ab and be their lines of action. The
position of the line of action ac of their resultant will now be de-
termined with respect to ab.
A
m
<- 5 ->
FIG. 43
By the principle of Art. 21, the moment of the resultant AC
(acting along ac in the space diagram) with respect to point n is
equal to the moment of its two concurrent components, AO and
OC. But the moment of AO is zero, so
Moment of AC = Moment of OC.
Also, the moment of OC with respect to n is equal to the moment
of its two concurrent components, OB and BC. Since the moment
of OB is zero,
Moment of OC = Moment of BC,
and Moment of AC = Moment of BC.
ACXr = BCXs.
Consider any point D, distant u from the line of action ab.
Since AC = AB + BC,
AC X u = AB X u + BC X u.
Add this equation to the one derived above. Then
AC(u + r) = AB X u + BC(u + s).
The Principle of Moments for parallel forces may now be stated :
The algebraic sum of the moments of two parallel forces with respect
to any point in their plane is equal to the moment of their resultant
with respect to the same point.
If the point D is taken on the line ac, the equation just derived
becomes
AB X r - BC X t = 0,
t AB
r~ BC'
26
APPLIED MECHANICS
[CHAP, ii
or, the perpendicular distances of the resultant from the forces are to
each other inversely as the forces.
Since by geometry three parallel lines divide any two inter-
secting straight lines in the same ratio, this principle holds true
for any diagonal distances, also.
The graphical application of this principle is often very advan-
tageous, so will be given here. If ab and be, Fig. 44, are the two
forces, join their lines of action by any
convenient straight line mn. From m lay
off ml on ab equal by scale to force be, and
from n lay off up in the opposite direction
on be equal by scale to force ab. Join pi,
cutting mn at o. The resultant passes
through o.
FIQ. 44
Problem 1.
cd
Show that- = -^f^ applies as well when AB and BC are in
T .DC/
opposite directions.
Problem 2. Check the position of the resultant as found in Problems of
Art. 23.
25. Resultant of Any Number of Parallel Forces, Graphically.
The graphical method of Art. 23 is readily extended to the case of
three or more parallel forces in a plane. In Fig. 45, AB is resolved
at any point m into A
AO and OB. OB is
combined with BC into a b
their resultant OC. OC
is combined with CD
into OD. OD is com-
bined with DE into OE.
AO and OE are now
the only forces left, so
they are combined into
the final resultant of the
system, AE, acting through point p.
follows :
(1) Draw the space diagram.
(2) Draw the force polygon, noting the resultant.
(3) Choose any convenient pole 0 and draw the rays.
(4) Parallel to the rays of the force diagram draw the corre-
sponding strings of the funicular polygon in the space diagram.
(a) Space Diagram
(b) Force Diagram
FIG. 45
The order of solution is as
ART. 26]
PARALLEL FORCES
27
(5) The intersection of the first and last strings determines the
position of the resultant.
If Bow's notation is used, each string has its corresponding ray
lettered similarly. For example, string ob is parallel to ray OB,
and is drawn between the two lines which enclose the "b" space,
ab and be. When the solution is begun, string oa is "free" until
intersected by the other free string to determine the line of action
of the resultant.
If the final point of the force polygon coincides with the initial
point so that the resultant R = 0 but the two "free" strings do
not coincide, the resultant of the system is a couple, as will be
discussed in Arts. 27-30.
In case the forces are not in the same plane, the graphical method
is used by finding the resultant of any two forces in their plane,
then the resultant of this with a third force in their separate plane,
r,
FIG. 46
9 Ibs. and the other forces as
and so on.
Problem 1. Fig. 46 represents a body 4 ft.
long with forces applied as shown. If F\ — 10
Ibs., Fz = 12 Ibs , F3 and F4 = 0 and F6 = 16 Ibs., i , , I , i ,
find the amount and position of the resultant. <"/ "5
. Ans. 38 Ibs., 2 ft. from the left end.
Problem 2. In Fig. 46, let F3 = 14 Ibs., F4
in Problem 1. Determine the resultant.
Ans. 15 Ibs. downward, 1.4 ft. from the left end.
Problem 3. Forces acting downward on the four vertical edges of a 3-ft.
cube are in order, 10 Ibs., 20 Ibs., 25 Ibs. and 15 Ibs. Combine the four forces
graphically.
Ans. R = 70 Ibs., 1.07 ft. from 20-25 face, 1.29 ft. from 15-25 face.
26. Resultant of Any Num-
ber of Parallel Forces, Alge-
braically. In amount and
direction the resultant R of
any number of parallel forces
is given by their algebraic sum.
The method of locating the
position of the resultant will
now be shown.
Let Flt F2, F3, etc., Fig. 47,
be any number of parallel forces and XOZ a plane of reference
normal to them. Consider first the forces ^i and F2 whose lines
of action pierce the plane of reference at B and D. Their result-
FIG. 47
28 APPLIED MECHANICS [CHAP, n
ant, FI + F2, lies in their plane and, by Art. 24, pierces the reference
plane at some point C such that
(Fi + Fz)AC = F! X I# + F2 X ZJD.
If this equation is multiplied by sin BAO, it becomes
If (Fi + F2) is combined with F3 in a similar manner, the result-
ing expression is
(F, + F2 + F3)z = FM + F2Z2 + F3z3.
By continuing until all of the forces of the system are combined
into their resultant R, the final relation is obtained :
Rx = FiXi + F2Z2 + F3z3 +, etc.,
R
In a similar manner the distance z of the resultant from axis OX
may be determined.
R
It will be noticed that the distance x of resultant R from the
axis OZ is the same as the distance x of any point in the resultant
from the plane ZOY. It follows, then, that the same moment
equation would be obtained for any axis in the ZOY plane parallel
to OZ. Since this is so, it is common to speak of the moment of a
force with respect to a plane parallel to it, meaning thereby the
moment of the force with respect to any axis in the given plane
normal to the force.
It will now be shown that the expression derived above for
locating the distance of the resultant from any axis in a plane
normal to the force system holds true as well for any inclined axis.
Let X'OZ be a plane at an angle 0 with the plane XOZ. Each
force, at its point of intersection with the plane X'OZ, may be
resolved into two rectangular components, F sin 0 in the plane
X'OZ, and F cos 6 normal to it. The components F sin 0 in the
plane X'OZ have no moment with respect to the axis OX', hence
the moment of the normal components is equal to the moment of
the original forces. From the preceding discussion the moment
equation for the normal components becomes
R cos 0 • z — FI cos 0 • zi + F2 cos 0 • 22 +, etc.
ART. 27] ' PARALLEL FORCES 29
If this equation is divided through by cos 6, it becomes identical
with that for axis OX or any axis in the plane XOY parallel to OX,
RZ = plZl + F& +, etc.
It must be remembered that z is the mutual perpendicular be-
tween the force and the inclined axis OX'.
Since the axis OX may have any position in the plane normal to
the forces, the axis OX' may have any position whatsoever; there-
fore the algebraic sum of the moments of any number of parallel forces
with respect to any axis is equal to the moment of their resultant with
respect to the same axis.
If the resultant R of a system of parallel forces is equal to zero,
but the moment of the system with respect to any axis is not equal
to zero, the system is equivalent to a couple, as will be discussed in
Arts. 27-30.
8000 10,000 16,000
Problem 1. Determine the position , / I , i » V ,
"^""3"" ">:'"
of] the resultant of the three downward
forces shown in Fig. 48.
Ans. 6.7 ft. from left end.
]L
Problem 2. Determine the position \ fyOQO 19000
of the resultant of the two upward FIGi 43
forces shown in Fig. 48.
Problem 3. A rectangular table 3 ft. wide and 4 ft. long has weights so
placed that the downward pressures on the four legs in order around the table
are as follows: A = 20 Ibs.; B = 26 Ibs.; C = 30 Ibs.; D = 24 Ibs. A
and B are at one end. Solve for the amount and position of the resultant.
Ans. R = 100 Ibs., 2.16 ft. from end AB, 1.32 ft. from side BC.
27. Moment of a Couple. Two parallel forces, equal in
amount, opposite in direction and with different lines of action,
constitute a couple, as FF, Fig. 49. The
perpendicular distance between them, /, is
called the arm of the couple. The product
of one force and the arm is called the
moment of the couple, or
Mam. = Ff.
The moment of a couple is the same with respect to any point in
its plane, as will be shown. Let 0 and 0' be any two points in the
plane of the couple, Fig. 49.
SM0 = FXOA+FXOB = FXAB = Ff.
Also, SM0' = F
30 APPLIED MECHANICS [CHAP, n
Since the resultant R of a couple is zero, moment is the only
effect of a couple. It follows, then, from the two preceding prin-
ciples, that a couple may be transferred to any place in its
plane or rotated through any angle and its effect will remain the
same.
Since moment is the only effect of a couple, it follows also that
any couple may be replaced by another of the same moment in
the same plane. Thus the rotary effect of a couple composed of
two forces of 8 Ibs. each, acting 3 ft. apart, is the same as that
of another in the same direction with forces of 4 Ibs. each, acting
6 ft. apart.
No single force can balance a couple. Since the resultant R of
the couple is zero, the resultant of the couple and another force
could not be zero.
A couple may be transferred to any plane parallel to its original
plane without change of effect. Since the moment of a couple with
respect to any point 0 in its plane is the same as its moment
with respect to an axis through 0, perpendicular to its plane, the
moment is independent of the location of the plane of the couple
along the axis. For example, if a steam pipe is being screwed into
a sleeve by means of two pipe wrenches (constituting a couple), the
effect is the same, no difference at what point along the pipe they
are applied.
28. Graphic Representation of a Couple. Since couples have
no properties but magnitude and direction, they may be repre-
sented graphically by vectors. The length of the vector repre-
sents to some scale the magnitude of the couple, and the direction
of the vector shows the direction of its plane and the direction of
its rotation. The vector is drawn perpendicular to the plane of
the couple. The convention commonly used with regard to the
arrow is that in which, if the couple is viewed from the head end
of the vector, the rotation of the couple appears
positive (counter-clockwise). Either of the
vectors, V, Vi, Fig. 50, 10 units long, vertical,
with the arrow pointing downward, represents
the moment ( — 10 f t.-lbs.) of either couple in
the horizontal planes as shown.
The position of the vector is immaterial, since
the moment of the couple is the same with
respect to any axis perpendicular to its plane.
ART. 29]
PARALLEL FORCES
31
29. Composition of Couples. The moment of the resultant
of any number of coplanar couples or of couples in parallel planes,
is equal to the algebraic sum of the moments of the component
couples.
By means of the principles of Art. 28, couples may be com-
pounded by combining their vectors. Since the position of the
vector is immaterial, the vectors of the couples may all be taken
through any given point, then added graphically. The resultant
vector represents completely the resultant couple.
Couples may also be combined directly. If the couples are in
the same plane (or in parallel planes), they may all be reduced to
couples having equal arms/,
with the forces parallel. If
these are superimposed, the
forces FI, Fz, F3j etc., com-
bine into their resultant F
in each case. Then
+£-*
f
In Fig. 51, couple Flffl
reduces to couple F3/. FlG- 51
Couples F2'/2 and F3'/3 reduce to F2f and F3/. Each set of forces,
Fij F2, F3, gives the resultant force F, so the resultant couple is F/.
If two couples to be combined are in intersecting planes, they
may be reduced to couples whose forces are equal each to each.
If the couples are then transferred, each in its own plane, so that
one force of each lies in
the intersection of the two
planes in opposite direc-
tions, as in Fig. 52, these
two forces neutralize each
B
ft
(a)
(b)
FIG. 52
other and may be removed
from the system. This
leaves the couple with forces FF and arm / in the plane ABCD.
If 0 is the angle of the two planes, and /i, /2 the arms of the
original couples after being transposed, the arm / is given by
Problem 1. In Fig. 52, transfer the couple whose arm is /i downward in
its plane until the upper force is at the intersection of the two planes and
determine the resultant couple.
Ans. Forces 2 F with arm ~
a
32
APPLIED MECHANICS
[CHAP, ii
FIG. 53
Problem 2. Couples of 10 Ibs. X 4 ft., 3 Ibs. X 16 ft., and 20 Ibs. X 5 ft.
are located in the same vertical plane, all rotating counter-clockwise. What
is their resultant couple? Ans. +188 ft.-lbs.
Problem 3. A couple of 10 Ibs. X 6 ft. in a vertical plane and one of
4 Ibs. X 20 ft. in a horizontal plane are to be combined. Both are negative,
viewed from the angle between the planes. Determine the amount of the
resultant couple and the slope of its plane.
Ans. -100 ft.-lbs. 6 = 36° 52' with the horizontal.
30. Resolution of a Force into a Force and a Couple. Any
force may be resolved into a force through a given point
and a couple. In Fig. 53, let F be the given force acting
at A, and let 0 be the given point. At 0 introduce
two opposite forces, FI and F2, each equal and paral-
lel to F. Since they neutralize each other, they do
not affect the system. Then FI and F constitute a
couple with moment Ff, and may be transferred to
any place in their plane, leaving force F2 equal to
the original force F, but acting at 0. joo
t\ -^i \
Problem 1. Resolve the forces of Fig. 54 into a force
at 0 and a couple. Each square is 1 ft. on each side.
Ans. R = 263.25 Ibs. 6 = 29° 26' with hor. Couple
= +329.3 ft.-lbs. 0
Problem 2. Resolve the forces acting upon the 3-ft. FIG. 54
cube in Fig. 55 into a force at 0 and a couple. Note: — Resolve each force
into its X, Y and Z components first.
Ans. R = 266 Ibs. a = 52° 15'. 0 = 127°.
7 = 121°. Couple = 902 ft.-lbs. Direction
of vector of couple is given by: a' = 85° 30';
0' = 46°; 7' = 136°.
31. Coplanar Parallel Forces in
Equilibrium, Algebraic Solution. In
general, any system of parallel forces
in a plane may be reduced to a single
force R and a couple with moment M.
If for any system both the resultant
R = 0 and the moment M = 0, the system is in equilibrium.
Conversely: // a system of coplanar parallel forces is in equilib-
rium, the resultant R = 0 and the moment M with respect to any
axis = 0.
If in a given system of forces which is known to be in equilibrium,
some of the forces are unknown, they may be determined by
FIG. 55
ART. 31]
PARALLEL FORCES
33
applying the conditions of equilibrium. The number of unknown
forces, however, must not exceed the number of independent equa-
tions which can be written, two in this case.
JQQQ 2000
-,t'\ A' L\ i'
" " f"
/<— ...... - 10- ......
FIG. 56
EXAMPLE.
A beam 10 ft. long, supported at the ends, carries three loads spaced as
shown in Fig. 56. What are the reactions at the
supports, #1 and R2, if the weight of the beam
itself is neglected?
Solution: — The beam is at rest, so the force
system acting upon it is in equilibrium. Appli-
cation of the equation 2F = R = 0, gives
Ri+R2- 9000 = 0.
Equation SM = 0, with any point on Ri as the center of moments, gives
10 #2 - 6000 X 2 - 1000 X 5 - 2000 X 9 = 0.
R2 = 3500 Ibs.
By substitution of the value of R2 in the equation above
Ri = 5500 Ibs.
As a check, an independent solution for Ri may be made by writing the
equation of moments with respect to any point on R2.
When the loading on a beam is symmetrical, no equations need be written,
since each reaction is one-half of the total load.
Problem 1. Solve for the reactions
Neglect the weight of the beam.
i and R2 of the beam shown in Fig. 57.
Ans. & = 1917 Ibs. Rz =^583 Ibs.
2,000
FIG. 58
Problem 2. Solve for the reactions Ri and R2 of the beam shown in Fig. 58.
(Note: — The weight of the beam, 2000 Ibs., is a uniformly distributed load
with its centroid at the middle of the beam.)
Ans. Ri = 15,580 Ibs. R2 = 11,420 Ibs.
Problem 3. Solve for the reactions Ri and R2 of the beam shown in Fig. 59.
Ans. Ri = 9833 Ibs. R2 = 3167 Ibs.
— -5— >\
-I
FIG. 59 FIG. 60
Problem 4. Solve for the reactions Ri and R2 of the overhanging beam
shown in Fig. 60. Ans. Ri = 40 Ibs. R2 = 1260 Ibs.
34
APPLIED MECHANICS
[CHAP, ii
32. Coplanar Parallel Forces in Equilibrium, Graphic Solu-
tion. The graphic relation corresponding to the principle of
Art. 31 is as follows:
// a system of coplanar parallel forces is in equilibrium, the force
polygon closes and the funicular polygon closes.
By the application of these two conditions of equilibrium, two
unknown forces may be determined.
EXAMPLE.
Determine the reactions Ri and R2 of the beam shown in Fig. 61.
Solution: — The space diagram, Fig. 61 (a), is drawn to scale. The force
polygon, Fig. 61 (b), is drawn to scale as far as known, AB, BC, CD. Since
A
4,500 1,500 7,000
a , \ b , \ c.
/
|<-4-~H<- 7'---->k-5-->
<-E>\
^r^^J^
tf^f- o e ~~~^—-~
*^4-*-f«n
(a} Space Diaqram
'** i i n -vi /
E A scales = 5,340 1 bs.
FIG. 61
(b) Force Diagram
Seal e-. I"-- 10,000 Ibs.
the system is in equilibrium, point A must be the closing point, but point E
is unknown. The location of point E is determined by the fact that the funic-
ular polygon must close also. Any convenient pole 0 is selected and the rays
OAt OB, OC and OD are drawn. The funicular polygon is begun at any con-
venient point on any one of the forces, as point m on force R\. String oa is
drawn parallel to ray OA across the "a" space. From the point at which oa
intersects ab, ob is drawn parallel to OB across the "6" space. Strings oc and
od are drawn in a similar way. Since the funicular polygon must close, string
oe must necessarily run from m to n. In the force diagram, ray OE must be
parallel to string oe in the space diagram, so point E is determined. Vector
DE represents the reaction Rz to scale and vector EA represents the reaction Ri.
Problem 1. A beam 13 ft. long, sup-
ported at the ends, has a load of 3000 Ibs.
4 ft. from the left end and one of 1000 Ibs.
3 ft. from the right end. Determine the
reactions.
Ans. R} = 2310 Ibs. R2 = 1690 Ibs.
Problem 2. Solve for the reactions Ri
200
L 1
'^<- 7- J
V " \
FIG. 62
and Rz of the overhanging beam shown in Fig. 62.
Ans. Ri = 544 Ibs.
-44 Ibs.
ART. 33] PARALLEL FORCES 35
Problem 3. Solve for the reactions Ri and R2 of the cantilever beam
shown in Fig. 63. ^Q
Ans. Ri = 1050 Ibs. R2 = 1200 Ibs. iff,
33. Equilibrium of Parallel Forces in ^ / 1, ~
Space. In case all of the forces of a *"/?2
given system of parallel forces are not in FlG- 63
the same plane, the principles of Arts. 31 and 32 must be slightly
modified.
Algebraically. // a system of parallel forces in space is in equi-
librium, the resultant R = 0 and the moment M = 0 with respect to
any axis in space.
Graphically. If a system of parallel forces in space is in equi-
librium, the projection of the system upon any plane constitutes a
coplanar system of parallel forces in equilibrium.
Some problems of this kind may be simplified by replacing one
or more pairs of forces by their resultant so as to reduce the system
to a coplanar system.
EXAMPLE.
A horizontal equilateral triangular plate ABC, 3 ft. on a side, is supported
at the vertices. What are the three reactions due to a load of 100 Ibs. acting
at a point on the median line 1 ft. from vertex A?
Solution: — The altitude of the triangle is 2.6 ft. The distance from the
base BC to the load is 1.6 ft. Equation SAf = 0 for axis through the edge BC
gives
100 X 1.6 = 2.6 A.
A = 61.5 Ibs.
By symmetry, reaction B = reaction C.
Equation ZF = 0 gives
B + C + 61,5 = 100
B + C = 38.5
B = C = 19.25 Ibs.
Problem 1. A stick of timber 12 ft. long and of uniform cross section is to
be carried by three men, one at the rear end, the others at the ends of a cross-
bar under the stick. How far from the front end should it be placed in order
that each man may carry the same weight. Ans. 3 ft.
Problem 2. A triangular flat plate ABC has side AB = 8 ft., BC = 6 ft.
and CA = 10 ft. If the plate is placed horizontally and supported at the
three corners, what are the three reactions due to a load of 100 Ibs. resting on
the plate at a point 1 ft. from side AB and 1 ft. from side BC!
Ans. A = 12.5 Ibs. B = 60.83 Ibs. C = 16.67 Ibs.
36
APPLIED MECHANICS
[CHAP, ii
GENERAL PROBLEMS.
Problem 1. A beam 4 ft. long weighing 40 Ibs. has a load of 100 Ibs. at
the left end and one of 30 Ibs. at the right. At what point will it balance?
Ans. 1.176 ft. from left end.
Problem 2. A force P of 40 Ibs. is acting vertically upward. Another
force Q of 20 Ibs. is acting vertically downward 2 ft. to the right of P. Deter-
mine the amount and position of their resultant.
Ans. 20 Ibs. upward, 2 ft. to the left of P.
Problem 3. A cantilever truss has dead loads and wind loads acting upon
it as shown in Fig. 64. Determine the amount and direction of the resultant
of these loads. Ans. 15,600 Ibs. at 7° 23' with the vertical, V.
FIG. 64
— 30
FIG. 65
Problem 4. Determine the amount, direction and position of the resultant
of the wind and dead loads on the truss shown in Fig. 65.
Ans. 21,920 Ibs. at 7° 30' with V. Resultant cuts lower chord 1.5 ft. to left
of middle.
Problem 5. Determine the amount, direction and position of the resultant
of the wind forces on the truss shown in Fig. 66.
Ans. 4670 Ibs. horizontal, 6.25 ft. below lower chord.
FIG. 66
Problem 6. If the total dead load of the truss shown in Fig. 66 is 12,000
Ibs., find the resultant of all the loads.
Ans. 12,875 Ibs. at 21° 15' with V, cutting lower chord 2.43 ft. to left of
middle.
Problem 7. A traction engine carries 15,000 Ibs. weight on the two driving
wheels and 4500 Ibs. on the steering wheels. The distance between the front
and rear axles is 11 ft. 8 in. Find the position of the resultant.
Ans. 2.69 ft. in front of drive wheels.
GENERAL PROBLEMS 37
Problems. A beam 6 ft. long has forces acting upon it as follows: 100 Ibs.
downward at the left end; 200 Ibs. upward 1 ft. from the left end; 400 Ibs.
downward at the middle; 300 Ibs. upward at the right end. Determine the
resultant. Ans. The resultant is a couple of + 800 ft.-lbs. moment.
Problem 9. Forces parallel to the Z axis and all acting in the same direction
are located as follows: 10 Ibs. at (2', 4'); 20 Ibs. at (!', 2'); 30 Ibs. at (4', 1').
Determine the amount and position of the resultant.
Ans. 60 Ibs. at (2.67', 1.83').
Problem 10. The sketch in Fig. 67 represents an A. T. & S. F. passenger
locomotive. Dimensions are given to the nearest foot. The weight on the
driving wheels is 147,400 Ibs.; on the truck is 28,600 Ibs.; on the trailers is
38,600 Ibs. The weight of the tender is 135,400 Ibs. Find the distance of the
front wheel from the edge of a turntable 100 ft. in diameter for perfect balance.
Ans. 18.2 ft.
o o o o
^5'4^--9 '--><-5*- -JO '- ->U- 8 '<- 7 - 5>U-7' <5 ^< 6>
FIG. 67
Problem 11. A force of 100 Ibs. is acting downward at each corner of a
horizontal equilateral triangle ABC, 10 ft. on each side. Determine the posi-
tion of the resultant. Ans. 5.77 ft. from each vertex.
Problem 12. A counter-clockwise couple of 2 Ibs. X 2 ft. in the XY plane
is to be combined with a counter-clockwise couple of 3 Ibs. X 1 ft. in the YZ
plane. Find the magnitude and direction of the resultant couple.
Ans. +5 ft.-lbs. Its plane is parallel to the Y axis and is at an angle of
36° 40' with the XY plane.
Problem 13. If a wagon is coupled too long for a platform scale to accommo-
date it, will the same result be obtained by weighing the front part and the rear
part separately and adding the weights?
Problem 14. Reverse the force at vertex C in Problem 11 and solve.
Ans. 100 Ibs., 8.66 ft. from AB.
Problem 16. Forces of 40 Ibs., 60 Ibs. and 80 Ibs. act vertically downward
at the corners A, B and C, respectively, of a horizontal equilateral triangle,
3 ft. on a side. Locate the resultant.
Ans. 1.156 ft. from AB; 0.578 ft. from BC.
Problem 16. Equal parallel forces act downward at the corners of a
horizontal triangle with sides of 6 ft., 8 ft. and 10 ft. Locate the resultant.
Ans. 2 ft. from the 8-ft. side; 2.67 ft. from the 6-ft. side.
Problem 17. A balance is slightly out of adjustment, one arm being longer
than the other. Will it give accurate results by weighing first in one pan, then
in the other, and averaging?
Problem 18. A workman closes a gate valve by exerting a pressure of
25 Ibs. with each hand at opposite sides on the rim of a hand wheel 2 ft. in
38 APPLIED MECHANICS [CHAP, n
diameter. At another time he thrusts a bar through the wheel and exerts the
pressure at only one side, 3 ft. out from the center. What pressure must he
exert? What is the difference in action in the two cases? Ans. 16| Ibs.
Problem 19. The movable weight of a steelyard weighs 6 Ibs. The short
arm is 3 ins. long. How far apart must the pound graduations be placed?
Ans. \ inch.
Problem 20. One end of a bar 10 ft. long is hinged at the wall and the
other rests on a smooth floor 6 ft. below. The bar carries a load of 200 Ibs.
at the middle. What are the reactions? (Note: — A smooth surface can exert
only a normal reaction.) Ans. 100 Ibs. each.
CHAPTER III.
NONCONCURRENT, NONPARALLEL FORCES.
34. Composition of Nonconcurrent, Nonparallel Forces in a
Plane; Graphic Methods. Let F,, F2, F3, etc., Fig. 68, be the
forces to be combined. FI and F2 are transferred along their lines
of action until they intersect at
m, where they are combined into
their resultant RI. RI is trans-
ferred along its line of action until
it intersects the line of action of
F3, at n. F3 is also transferred
along its line of action until it is
in the proper position, F3', where
it is combined with RI into the resultant of all three forces R2. If
there are other forces the same procedure may be followed until
the final resultant is found.
If the forces are so nearly parallel that no intersection can be
obtained, any one of the forces may be broken up into two com-
ponents. Combination with the other forces may now be made as
c\d
FIG. 68
(a)
Force Diagram Space Diagram
FIG. 69
above. As in the case of parallel forces, both space and force
diagrams should be used. In Fig. 69, let the forces be AB, BC
and CD, acting along ab, be and cd respectively. Aim, AB may be
resolved into two components, AO and OB. OB and BC may be
combined into their resultant OC at n. OC and CD may be com-
39
40
APPLIED MECHANICS
[CHAP, in
bined into their resultant OD at p. Finally, AO and OD may be
combined into their resultant AD at q. AD is then the resultant
of the entire system.
Another force, equal and opposite to the final resultant and
colinear with it, will hold the system in equilibrium.
Also, conversely, // a system of nonconcurrent forces in a plane
is in equilibrium, the force polygon closes and the funicular polygon
closes,
If the force polygon closes but the funicular polygon does not
close, the resultant is a couple. In Fig. 70, force vectors AB, BC
and CD form a closed polygon,
with point D coinciding with
point A. These forces act along
lines of action ab, be and cd re-
spectively in the space diagram.
From any point 0, rays OA, OB,
OC and OD are drawn. In the
space diagram the corresponding
strings oa, ob, oc and od are drawn.
Strings oa and od are parallel but not colinear, so the system is
reduced to the two equal parallel forces AO and OD acting /
distance apart.
Problem 1. Combine the wind and
dead load forces acting upon the truss
in Fig. 71 into their resultant, graph-
ically
Ans 27,540 Ibs. at 4° 10' with V.
R cuts AB 0.63 ft. to the left of the
middle.
35. Composition of Nonconcurrent, Nonparallel Forces in a
Plane; Algebraic Method. The amount of the resultant R of
any system of nonconcurrent, nonparallel forces in the same plane
is given by the equation
R =
(a)
Force Diagram
Space Diagram
FIG. 70
Its direction angle 0 with the x axis is given by
As in the case of parallel forces, the theorem of moments is
readily extended to this case:
ART. 36] NONCONCURRENT, NONPARALLEL FORCES 41
For any system of forces in a plane, the moment of the resultant of
the system with respect to any point is equal to the algebraic sum of the
moments of the several forces with respect to the same point.
By means of this principle the position of the resultant R may be
determined by writing the equation of moments with respect to
any point. If the moment arm of the resultant is denoted by a,
and the moment arms of the several forces by ai, 02, etc.,
Ra = Fidi + F202 + , etc.
Another force equal and opposite to R and colinear with it will
"hold the system in equilibrium.
EXAMPLE.
As an example, Problem 1 of Art. 34 will be solved by the method of this
article. The vertical components of the wind forces in turn are 866, 1732 and
866. The horizontal components are in turn 500, 1000 and 500.
2FX = 500 + 1000 + 500 = 2000.
ZFy = 866 + 1732 + 866 + 3000 + 6000 + 6000 + 6000 + 3000 = 27,464.
)* = 27,540 Ibs.
R = (2FX)2 + (SP
The angle 0 with the horizontal is given by
2FV _ 27,464
-"" =
0 = 85° 50'.
To locate R, use point A as the center of
moments.
27,540 X a = 24,000 X 15 + 4000 X 8.66.
a = 14.33ft.
Problem 1. In Fig. 72, determine the
resultant of the three forces in amount and
direction. Determine also its perpendicular
distance a from point A.
Am. R = 3228 Ibs. 0 = 85° 05' with K.
a = 9.54 ft.
36. Reduction of a System of Forces
to a Force and a Couple. Any system of
forces in a plane may be reduced to a force
through any given point and a couple.
Let F, FI, etc., Fig. 73, be the system of
forces. At any point along its line of
action, F may be resolved into its X and
Y components, Fx and Fy. By Art. 30,
force Fy may be resolved further into an
equal force Fy' through 0, and a couple Fy X OM. Fx may also
FIG. 73
42
APPLIED MECHANICS
[CHAP, m
be resolved into an equal force through 0, and a couple Fx X ON.
Each force in turn may be resolved similarly. The resultant of
the forces at 0 is
R
200
and the resultant of all the couples is a single couple ZM.
For equilibrium of the system,
R = 0 and 2M = 0.
Conversely, // a system of nonconcurrent,
nonparallel forces in a plane is in equilibrium,
the algebraic sum of the forces along any line
is zero and the moment with respect to any axis
is zero.
K 2' >1 Problem 1. Reduce the force system shown in Fig.
FIG. 74 74 to a force through point 0 and a couple.
Ans. R = 527 Ibs. M = -687 ft.-lbs.
37. Two-force Members, Three-force Members, Etc. The
student has doubtless already noticed that some members of
structures have forces applied at only two points as for instance
any member of a common bridge truss. (In this case, only the
actions of the other members upon the member considered are
taken into account, the weight of the member itself being ne-
glected.) Such members of structures are called Two-force Mem-
bers. In the crane shown in Fig. 75, the brace BE is a two-force
FIG. 75 FIG. 76
member. Its free body diagram is shown in Fig. 76(a), in which
the vectors at B and E represent the actions of the post and the
boom respectively upon the free body. The forces acting upon
ART. 37] NONCONCURRENT, NONPARALLEL FORCES 43
a two-force member are necessarily axial, producing direct com-
pression or tension in the member, for if they were not axial they
would not be colinear and could not hold the member in equi-
librium. Since the internal stress in a two-force member is
always axial, a section may be made through it in taking a free
body.
If any member of a structure has forces applied at three or more
points, it is called a Three-force Member. Since the stress is in
general not axial, a section should not be made through a three-
force member in taking a free body, but the entire member should
be taken. If another three-force member joins the one considered
as the free body, the X and Y components of the action of the
other member upon the free body must be introduced if the direc-
tion of their resultant is not already known.
If a body is in equilibrium under the action of three forces, they
must intersect in a common point or be parallel, and the resultant
of any two must be equal and opposite to the third. The boom
CEF, Fig. 75, is a three-force member, shown as a free body in
Fig. 76 (b). The force at F is the external load P. That at E is
equal and opposite to the force at E on the brace EB. Since these
two are known in direction, their intersection at G determines
another point in the line of action of the force at C. Likewise, if
the entire crane is considered as a free body, the direction of the
reaction at A is determined by the intersection of the lines of
action of force P and the horizontal reaction at D.
If a body is in equilibrium under the action of four forces, the
resultant of any two must necessarily be equal and opposite to the
resultant of the other two. If one of the four forces is wholly
known and the directions of the others are known, the three
unknown forces may be determined. This is illustrated in Fig.
76 (c). The resultant of the forces at C and D must be equal and
opposite to the resultant of the forces at A and B; that is, R is
equal and opposite to RI and is colinear with it.
The method of procedure in solution is to intersect the forces in
pairs and join the two points of intersection. This gives the line
of action of the resultant of each pair. With the force C known of
the system C, D, R, the two unknown forces D and R may be
determined. Resultant R is equal and opposite to R1} so the
system Ri, A, B, can be solved.
If five or more forces are acting upon a body in equilibrium, two
44
APPLIED MECHANICS
[CHAP, in
or more can usually be combined so as to reduce the system to a
four-force system, after which it can be solved as above.
38. Problems in Equilibrium; Forces in a Plane. The prin-
ciples of Arts. 34-37 are useful in the solution of an important
class of problems occurring in engineering work. The usual case
is that in which a certain system of forces is known to be in equi-
librium but some of the forces are unknown and are to be deter-
mined.
EXAMPLE 1.
The cantilever truss of Fig. 77 (a) is loaded with roof or "dead" loads as
shown. Determine the reactions at E and C by the algebraic method.
FIG. 77
Solution. — The reaction at E is the same as the stress in the strut or com-
pression member DE. Since DE is a two-force member, its stress must be
axial) so is known in direction. The reaction at C is unknown both in amount
and direction, but must necessarily pass through point C. The free body
diagram is shown in Fig. 77 (b). Let the reaction at C be called P and let that
at D be called Q. Also let the unknown angle between P and the vertical be
called 6. Since the truss is in equilibrium,
2FX = 0, 2Fy = 0 and SM = 0.
The first equation gives
Q X 0.866 - P sin 0 = 0.
The second equation gives
Q X 0.5 + P cos0 - 5000 = 0.
The third, with C as center of moments, gives
1250 X 17.32 + 2500 X 8.66 = Q X 11.5.
Q = 3750 Ibs.
By substitution of the value of Q in the equations above,
Psin0 = 3250.
Pcose = 3125.
By division, tan 6 = 1.04.
6 = 46° 08'.
Sin 46° 08' = 0.721.
P =
3250 = 3250
sin0 ~ 0.721
= 4510 Ibs.
ART. 38] N ON CON CURRENT, NONPARALLEL FORCES 45
P sin 0 is really the X component of force P and P cos 0 is the Y component,
so the solution would be practically the same if the unknown force P were
replaced by Px and Py. Then
p = VpTTTy.
With Px and Py replacing P, a simpler solution for the three unknown forces
is made possible. By writing the equation ZAf = 0 with respect to the three
points of intersection of the unknown forces in turn, only one unknown quantity
occurs in each equation. Elimination between two equations is thus avoided.
This method should be used if all distances are known or are easily obtained.
EXAMPLE 2.
Determine all the internal stresses in the members of the pin connected
bridge truss shown in Fig. 78.
Solution: — The truss is symmetrical
and the loading is symmetrical, so each
reaction is one-half of the total load of
10,000 Ibs.
R, = R2 = 5000 Ibs.
All of the members are two-force mem- ft
bers, so sections may be made through
them as desired in taking a free body.
Let a section be made through the
FIG.
truss at ran, .and let the part at A be taken as the free body, as shown in Fig.
79 (a). The force FI is the internal stress in A B, acting now as an external
force on the free body, and must be compression in order to balance R\. Simi-
(a)
T
(c)
larly the force F2 is the internal stress in AC and must be tension in order to
balance FI. Since this free body is in equilibrium, SFV = 0 and SFX = 0.
Equation SFj, = 0 gives
5000 - F! sin 45° = 0.
FI = 7070 Ibs. compression.
Equation 2FZ = 0 gives
F, - 7070 cos 45° = 0.
F2 = 5000 Ibs. tension.
The next free body taken is the joint at B, enclosed by section pq. The
free body diagram is shown in Fig. 79 (b). There are two known forces acting
on the free body, the load, 2000 Ibs., and the stress in AB. The action of this
46 APPLIED MECHANICS [CHAP, m
force on B must be equal in amount and opposite in direction to the force which
AB exerts on A, so is 7070 Ibs. acting as shown. Inspection of the known
vertical forces acting at B shows that there is a larger force upward than down-
ward. Therefore for equilibrium F4 must have a component downward, so is
tension as shown. For equilibrium horizontally, F3 must be compression.
Equation 2Fy = 0 gives
7070 sin 45° - 2000 - F4 sin 45° = 0.
F4 = 4240 Ibs. tension.
Equation 2FX = 0 gives
7070 cos 45° + 4240 cos 45° - F3 = 0.
Fs = 8000 Ibs. compression.
It is not necessary that the true direction of the unknown stresses be deter-
mined before solution as above. If in either of the force diagrams the direc-
tion of an unknown force had been assumed incorrectly, the value obtained
would have been the same numerically but negative in sign.
Since the truss and loading are both symmetrical, the stresses in corre-
sponding members on the two sides of the truss are equal, so the solution need
not be carried further unless it is desired to complete it as a check.
If the stress in only one member had been required as for instance that in
BD, a shorter method would have been as follows: Let the section rs, Fig. 78,
be passed through the truss and let all to the left of the section be taken as the
free body. Fig. 79 (c) shows the free body diagram. There are now three
unknown forces, but they are not concurrent, so the problem can be solved.
Equation SMc = 0 gives
Fs X 10 - 5000 X 20 + 2000 X 10 = 0.
F3 = 8000 Ibs. compression.
Stresses F2 and F4 can now be determined if desired.
Equation 2Fy = 0 gives
F4 sin 45° + 2000 - 5000 = 0.
F4 = 4240 Ibs.
Equation Sl/B = 0 gives
F2 X 10 - 5000 X 10 = 0.
F2 = 5000 Ibs.
The graphic method is especially well adapted to the solution of problems
of this kind. Each joint in turn is taken as a free body and the unknown
forces are determined by the principle that the force polygon for the free body
in equilibrium must close.
In Fig. 80 (a) is shown the truss of Fig. 78 with the same loading but with
Bow's notation of lettering. The line of action of the 6000-lb. load is produced
backward so that all of the loads are taken in order between the reactions.
The force polygon for the first joint at the left, eaf, is shown in Fig. 80(b),
EA being the known reaction. AF = 7070 Ibs. and FE = 5000 Ibs. are
determined by this polygon, and are the stresses in af and fe respectively.
Since the force polygon must close, with the vectors following each other
around the polygon, the arrows must be in the direction shown. The stress
ART. 38] NONCONCURRENT, NONPARALLEL FORCES 47
AF is therefore downward toward the joint, compression, while the stress FE
is away from the joint, tension. The sequence of the letters in the direction
in which the polygon was drawn also indicates the direction of the stress. The
arrows should be placed on the members in the space diagram as the direction
of each stress is determined. The joint fahg is the next free body, the two
forces FA and AB being known. In Fig. 80(c) these are laid down in order
and by the closing of the polygon, forces BG and GF are determined. It should
be noted that in drawing the force polygon all of the known forces must be
taken first. Also, less confusion results if the forces are always taken in the
same order around the joints. The clockwise direction will usually be used,
as in the preceding solution.
Fig. 80 (d) shows the force polygon for the middle joint, HI and IE being
the unknown forces determined. In this polygon point 7 coincides with point
F and vectors EF and IE are superimposed. Fig. 80(e) shows the force
polygon for joint iticd, force ID being determined. This completes the solu-
tion, but the force polygon for the last joint may be drawn as a check, Fig.
80(f).
Since each of the internal stress vectors occurs twice, the force polygons
may be superimposed as constructed, so as to form one complete diagram
Fig. 80(g). The arrows on the stress
diagram may be omitted but they
should always be placed on the mem-
bers in the space diagram.
EXAMPLE 3.
The A-frame shown in Fig. 81 (a)
supports a load of 8000 Ibs. at the mid-
dle of member BD. Determine the
pin reactions at B, C and D caused by
this load, if the floor is considered to
be smooth. Use the algebraic method.
Solution: — Since the floor is smooth,
the reactions at A and E are neces-
sarily vertical. The frame and load-
ing are symmetrical, so each reaction
•
is 4000 Ibs. If either the frame or the loading were not symmetrical, the
moment equations for the entire frame as a free body would determine the
48 APPLIED MECHANICS [CHAP, m
reactions. The structure consists entirely of three-force members, so each
must be taken as a free body. The cross bar BD, Fig. 81 (d), is considered
first. The known force is 8000 Ibs. downward at the middle. Since a three-
force member joins it at B and another at D, the vertical and horizontal com-
ponents of the reactions must be used. Either by moments or by symmetry
Bv = Dv = 4000 Ibs.
Equation 2FX = 0 gives BH = DH, but neither one can be evaluated from
this free body. Member AC is next taken as the free body. Forces A and
Bv are now known; forces BH, CH and Cv are unknown.
Equation 2FV = 0 gives
4000 - 4000 - Cv = 0.
Cv = 0.
Equation ZA/c = 0 gives
BH X 6 + 4000 X 6 - 4000 X 9 = 0.
BH = 2000 Ibs.
Since BH = DH, DH = 2000 Ibs.
Equation ZFX = 0 gives
BH = CH = 2000 Ibs.
The pin reaction at C is therefore 2000 Ibs. horizontal.
The pin reactions at B and D are given by
B = D = V20002 + 40002 = 4472 Ibs.
The angle 6 with the horizontal is given by
4000
6 = 63° 28'.
Problem 1. Solve for the stresses in the members of the truss of Example 1.
Ans. AB = 2500 Ibs. T. BC = 3750 Ibs. T. AD = 2165 Ibs. C. BD =
2165 Ibs. C. DC = 0.
Problem 2. Consider the load at D, Fig. 78, to be changed to 4000 Ibs. and
solve for all stresses in the members.
Ans. AB = 7780 Ibs. C. AC = 5500 Ibs. T. BC = 4950
Ibs. T. BD = 9000 Ibs. C. CD = 3540 Ibs. T. CE = 6500
Ibs. T. DE = 9200 Ibs. C.
Problem 3. The supporting cross bar AB of a platform
is 6 ft. long and holds weights as shown in Fig. 82. The
inner end is fastened to the wall by a hinge at A. A cable
BC at an angle of 30° with the vertical supports the other
end. Solve for the tension in the cable BC and the reac-
tion at A in amount and direction.
Ans. BC = 846 Ibs. R = 630 Ibs. 6 = 47° 50' with H.
800 400 Problem 4. Consider each member of the A-frame in
FIG 82 Example 3 to weigh 100 Ibs. per linear foot and determine
the pin reactions B, C and D.
Ans. C = 3255 Ibs. horizontal. B = D = 5635 Ibs. 6 = 54° 42' with H.
ART. 39] NONCONCURRENT, NONPARALLEL FORCES 49
39. Composition of Nonconcurrent, Nonparallel Forces in
Space. A system of forces, F, F\, F2, etc., not in the same plane,
may be reduced to a single force through any given point and a
couple. Let F, Fig. 83, be one of the forces of the system. In-
troduce at point 0 the opposite forces F' and F", each equal and
parallel to F. This does not affect the system in any way, since
they neutralize each other. Then F and F" constitute a couple
and the remaining force is F' equal to F, acting through point 0.
XX
I
FIG. 83 FIG. 84
This may be repeated with each force in turn. Each force of the
concurrent system at 0 is then resolved into its X, Y and Z com-
ponents and these are recombined, giving
R =
as shown in Fig. 84. The direction cosines are given by
S^v
cos/3 = -;
The couples may be combined by means of their vectors,
exactly the same as the force vectors were combined above but
a simpler method is the following. At any point along its line
of action, each one of the original forces may be resolved into its
X, Y and Z components. The algebraic sum of the moments of
these components with respect to the X, Y and Z axes in turn
gives Mx, My and Mz, which may be represented by their vectors.
The vector of the resultant couple is given by
M= VMX2 + My2 + Mz\
Also,
Mx My Mz
cos ai =-- -j£ ; cos fa = -^ ; cos Oi = -^ •
In general, then, a system of this kind tends to produce a trans-
lation of the body acted upon in the direction of R, and a rotation
in the plane of the resultant couple M.
50
APPLIED MECHANICS
[CHAP, in
X = 0, XFy = 0, SF, = 0, SMX = 0, SMy = 0 and SM2 = 0,
the force system is in equilibrium.
Conversely, // a body acted upon by a force
system of this kind is in equilibrium, these six
conditions are true.
Problem 1. Combine the forces shown in Fig,
85 into a force at 0 and a couple. The sides of the
cube are 2 ft. long.
Ans. R = 150 Ibs. a = 67° 50'. (3 = 153° 30'.
6 = 76° 13'. M = 305 ft.-lbs. ai = 71° 40'.
FIG. 85 ft = 90°. 0i = 161° 40'.
40. Problems in Equilibrium; Forces in Space. The condi-
tions of equilibrium for nonconcurrent, nonparallel systems of
forces in space as given in Art. 39 may be taken in sets of three,
as follows: ZFX = 0, SFtf = 0, 2MZ = 0.
2FX = 0, ZF* = 0, I>My = 0.
2Fy = 0, 2F, = 0, ZMX = 0.
It will be seen that the first set gives the conditions necessary
for equilibrium of a coplanar system of forces in the X Y plane; the
second set gives the conditions necessary for equilibrium of a
coplanar system of forces in the XZ plane; the third set gives the
conditions necessary for equilibrium of a coplanar system of forces
in the YZ plane. Also, 2FX and 2Fy are the X and Y components
of the projections of the forces on the XY plane and 2MZ is the
moment of these forces in that plane. Then it follows:
// a syst m of nonconcurrent, nonparallel forces in space is in equi-
librium, the proj ctions of these forces on any plane constitute a system
of forces in equilibrium.
By means of these prin-
ciples, unknown forces not
to exceed the number of
equations may be deter-
mined in any system which
is known to be in equi-
librium.
Side View
FIG. 86
End View
EXAMPLE 1.
Fig. 86 shows three views of a
simple windlass. It is required to determine P, R' and R" for' the position
shown, R' and R" being the reactions at A and B respectively.
ART. 40] NONCONCURRENT,NONPARALLEL FORCES 51
Solution: — The free body diagram for the top view, the projection of the
figure on the XZ plane, is shown in Fig. 87 (a). That for the side view, the
projection on the XY plane, is shown in Fig. 87 (b). That for the end view,
FIG. 87
the projection on the YZ plane, is shown in Fig. 87 (c). In these diagrams R'
is replaced by its horizontal and vertical components, RH and Rv '. Also R"
is replaced by its horizontal and vertical components, RH" and Rv". The
equation SM<? = 0 for Fig. 87 (c) gives
P X 18 = 300 X 3.
P = 50 Ibs.
The four other unknown forces in this projection cannot be determined.
With P known, the unknown forces in Fig. 87 (a) can now be deterndned.
PH = P cos 45° = 35.35 Ibs.
Equation 2MA = 0 gives
35.35 X 6 = RH" X 5.
RH" = 42.42 Ibs.
Equation SMs = 0 gives
35.35 X 1 = RH' X 5.
RH' = 7.07 Ibs.
In Fig. 87 (b), Pv = P sin 45° = 35.35 Ibs.
Equation SAfu = 0 gives
35.35 X 6 + Rv" X 5 = 300 X 3.
Rv" = 137.6 Ibs.
Equation SAf a = 0 gives
35.35 X 1 + 300 X 2 = Rv' X 5.
Rv' = 127.1 Ibs.
The amount and direction of R' and R" may now be determined if desired.
EXAMPLE 2.
Fig. 88 (a) shows the dimensions, position and loading of a derrick. Deter-
mine the external reactions due to a load of 1200 Ibs.
Solution: — The external forces on the derrick constitute a nonconcurrent,
nonparallel system in space, but by taking different parts of the derrick as free
bodies in turn, only concurrent systems need be considered. The first free
52
APPLIED MECHANICS
[CHAP, m
body to be considered is the pulley at A, shown in its free body diagram in Fig.
88 (b). If friction is neglected, the tension in any cable is constant throughout
its length. Then equation 2FV = 0 gives
4 T = 1200 (neglecting the slight angularity).
T = 300 Ibs.
FIG. 89
The pin at B has forces acting upon it as shown in Fig. 89 (a), forces 7\ and
P being unknown. Fig. 89 (b) shows the graphic solution, from which T\
scales 268 Ibs. and P scales 1740 Ibs.
Consider next a section made by the horizontal plane X — X, Fig. 88 (a),
and let the part above the plane be taken as the free body. The system of
forces acting on this free body is concurrent, but not coplanar. The horizon-
tal projections of these forces, however, constitute a coplanar, concurrent
system in equilibrium and so may readily be solved. From Fig. 89 (b) the
horizontal component of 5 T{ is 1250 Ibs. From Fig. 90 (a) and (b) it is seen
that the horizontal components of the stresses in CD and CE are each 1250
Ibs. The stress in each, as shown in Fig. 90 (c), is 1767 Ibs.
1,250
(a}
FIG. 90
FIG. 91
The compression in the mast is determined by considering the vertical
forces on the free body above plane X — X , Fig. 88 (a). Let the compressive
stress in the mast be called V. The vertical component of the stress in CB is
480 Ibs., as shown in Fig. 89 (b). The vertical component of the stress in CD
is 1250 Ibs. The vertical component of the stress in CE is 1250 Ibs. as shown
in Fig. 90(c). Equation 2FV = 0 gives
V - 1250 - 1250 - 268 - 480 = 0.
V = 3248 Ibs.
ART. 41] NONCONCURRENT, NONPARALLEL FORCES 53
Next take the socket at F as the free body, Fig. 91 (a). Equation 2Fy = 0
gives
#F = 3700 Ibs.
Equation 2FX = 0 gives
Rn = 1818 Ibs.
R = VRH* + Rv2 = 4120 Ibs.
1818
The angle with the vertical, 9 = tan"1 = 26° 10'. See Fig. 91 (b).
If framed members are inserted at DF and EF, they will carry the horizontal
components of the stresses in CD and CE to the foot of the mast.
Problem 1. Determine the reactions and the force P on the windlass of
Example 1 above, (1) when the handle is horizontal; (2) when the handle is
vertical.
Ans. (1) #F' = 1301bs. RH' = 0. #F" = 1201bs. RH" = 0. P = 501bs.
(2) Rv' = 120 Ibs. RH' = 10 Ibs. Rv" = 180 Ibs. RH" = 60 Ibs.
P = 50 Ibs.
Problem 2. In Example 2 above, consider the whole derrick as the free
body and solve for the vertical and horizontal components of the reaction at
the foot of the mast, and for the stresses in the legs.
Problem 3. Consider the boom of the derrick in Example 2 above to be
lowered until it is in the horizontal position. Determine the value of angle a,
(1) for maximum tension in EC; (2) for maximum compression in EC. (Do
not consider that the boom works in the smaller angle DFE, although it may
be so used.) Solve for the stresses in the boom, the mast and the stiff legs for
each position.
Ans. (1) 1740 Ibs. C. 4065 Ibs. C. 2350 Ibs. T. in EC. 1 175 Ibs. T. in DC.
(2) 1740 Ibs. C. 135 Ibs. C. 2040 Ibs. C. in EC. 0 in DC.
41. Cord Loaded Uniformly Horizontally. A flexible cord
suspended from two points forms a smooth curve. Two cases will
be considered: First, that in which the cord carries a load which
is uniformly distributed horizontally. Second, that in which the
cord carries a load which is uniformly distributed along the cord.
The second case will be discussed in Art. 42.
Fig. 92 represents a part of a cord carry-
ing a load uniformly distributed horizon-
tally. Let w be the weight carried per
horizontal unit. Let 0 be the lowest
point on the cord, B any other point,
H the tension at 0, P the tension at B,
and x the horizontal distance between 0 and B. The total load
x
on length OB is then wx, acting at ^ distance from 0. The part
Zi
OB is in equilibrium under the action of the three forces, H, P and
54 APPLIED MECHANICS [CHAP, m
wx, which are therefore concurrent at A. The equation SM^ = 0
gives wx* = 2 Hy.
»
FIG. 93
This is the equation of a parabola with its origin at 0 and its
axis vertical. If Z is the total span and d is the sag at the middle,
Fig- 93,
ir _
~8d'
If T7 is the tension at the support, equations ZFy = 0 and
= 0 give
= H.
By squaring the two preceding equations, adding, and extracting
the square root,
Also, from the figure,
„
tan 0! = —
Let s represent the length of the cord between the points of
suspension. The value of s is given by a logarithmic equation
derived in the calculus. This equation is accurate but is cum-
bersome to use. In order to obtain a simpler expression the log-
arithmic equation is expanded into a converging series, of which
the third and succeeding terms are so small that they may be
neglected without appreciable error. This gives
In terms of I and d,
, /A
s = l+-. (Approx.)
ART. 41] NONCONCURRENT, NOJSTPARALLEL FORCES 55
If s and H are given to find I, a cubic equation results. The second
term is comparatively small, however, so that I may be replaced
by s in it, giving
A tightly stretched horizontal wire very closely approximates
the condition of uniform loading horizontally, as do also the cables
of a suspension bridge, Fig. 94, since the extra weight of the cables
Fia. 94
and hangers toward the ends is a small part of the total load carried.
The lengths -of the hangers in the so-called "catenary" trolley wire
construction, Fig. 95, are really computed by the parabolic formu-
las of this article and not by the true catenary formulas. Allow-
ance must be made for the elasticity of the material used.
Insulated Support
1 l^r
— r— ^Messenger
p Y —
Cnble.
orHH
Hangers
FIG.
95
7?<?//e/ M^?
For a wire with a sag of 1 per cent of the span, the error in H, T
or d, compared with the value given by the correct catenary, is
about A: of 1 per cent. For one with a sag of 10 per cent of the
span, the error is about 2 per cent.
Problem 1. A steel wire weighing 0.04 Ib. per foot has a span of 200 ft.
and a tension at the lowest point of 300 Ibs. What is the sag? What is the
length of the wire? What is the amount and direction of the tension at the
supports? Ans. 8 in. 200.00592 ft. 300.027 Ibs. 0=0° 46'.
Problem 2. The cables of a suspension bridge have a span of 1200 ft., carry
a load of 800 Ibs. per linear foot per cable, and have a sag of 40 ft. at the middle.
Determine the tension at the middle, the tension at the ends and the length of
the cables. Ans. H = 3,600,000 Ibs. T = 3,631,900 Ibs. s = 1203.56 ft.
Problem 3. A messenger cable for a "catenary" trolley system weighs
0.3 Ib. per foot and is stretched between supports 100 ft. apart with a tension
of 2300 Ibs. What is the sag? (Assume H = T.) Ans. 0.163 ft.
56
APPLIED MECHANICS
[CHAP, m
42. The Catenary. When the load on a cord is uniformly
distributed along its length, the curve which the cord assumes is
called the catenary. The equation of
the catenary curve will now be derived.
Let w be the weight of the cord per
unit length. Let 0, Fig. 96, be the
lowest point on the cord, A any other
point, s the length of the cord from
0 to A, H the tension at 0 and T the
tension at A . Also let H be represent-
ed by the weight of an imaginary
length of the cord c, or H = we. In
Fig. 96 (a) the length of cord s is shown
as a free body in equilibrium. In Fig. 96 (b) is shown the force
triangle, from which the relation is obtained,
ws
- = tan 0,
we
FIG. 96
or
dx c
dy2 = ds"2 - dx*,
By squaring and solving for dx,
dx =
cds
or
By integration
(1)
The quantity e is the base of the Naperian system of logarithms,
and its numerical value is 2.718,28. The reduction to common
logarithms is made by the relation
0.4343 log* A = logio A.
Reduced to exponential form, equation (1) becomes
ART. 42] NONCONCURRENT, NONPARALLEL FORCES 57
Solution for s gives
If this value of s is substituted in the original equation, there is
obtained the expression
( - ~-\
dy = \ \e° — e c)dx.
If the origin is at 0 and dy is integrated between the limits 0
and y, a complicated expression results. A simpler expression is
obtained by using 0' as the origin. The integration of dy is then
between the limits c and y.
CVdy = £( f±e**dx - r±e~~c
Jc 2\J0 c Jo c
By integration,
/ x _x\
y - c = \ec + e c ) - c,
cf *
y = \ec
or y = 2\ec + e C- (3)
By squaring (2) and (3) and subtracting,
2/2 = s2 + c2. (4)
From (1) and (4)
x = cloges-±Jl. (5)
c
From the relation of the sides of the force triangle, Fig. 96 (b),
T* = w2c2 + wV = w2(s2 + c2) = w*y2.
T = wy. (6)
The related quantities are as follows:
Length Unit Weight Tension Span Deflection
2s w T 2xi = I y - c
The most useful problems, those in which the length, span and
weight, or the deflection, span and weight are given, can be solved
for the unknown quantities only by trial, on account of the form
of the logarithmic equation.
EXAMPLE 1
A cable 800 feet long, weighing 2 Ibs. per foot, is stretched between two
points on the same level with a tension of 1200 Ibs. What is the sag and the
span?
58 APPLIED MECHANICS [CHAP, m
Solution: — w = 2 and T = 1200. Equation (6) gives
y = 600 ft.
From equation (4)
c = 447.2 ft.
The sag is y - c = 152.8 ft.
From equation (5) the span is 2 x = 2 X 447.2 loge Tryv
2x = 719.8 ft.
EXAMPLE 2
If a cable weighing 2 Ibs. per foot is stretched between points 800 feet apart
and sags 100 feet, what is the tension and the length of the cable required?
Solution: — This problem can be solved only by trial.
y=c + 100.
From equation (4)
c2 + 200 c + 10,000 = s2 + c2.
s = 10 V2 c + 100.
From equation (5)
400 = o log. 10 ^27 + 100 + 0 + 100.
It is found by trial that c = 810 will nearly satisfy this equation. With
this value of c,
y = 910 and T = 1820.
From equation (4)
S2 = y2 _ <*
s = 414.7 ft.
Total length = 2 s = 829.5 ft.
Problem 1. A wire 300 feet long, weighing 0.01 Ib. per foot, has a tension
of 4 Ibs. at each end. What is the span and the sag?
Ans. 292.7 ft. 29.2 ft.
Problem 2. A 1-inch cable, weighing 1.58 Ibs. per foot, carries telephone
cables and supporting cross pieces weighing 0.22 Ib. per foot. The span
between towers is 862 feet and the sag is 50 feet. Find the length of the cable
and the tension. Ans. 864.2 ft. 3408 Ibs.
Problem 3. A chain 50 feet long, weighing 3 Ibs. per foot, is stretched
between two points on the same level 40 feet apart. What is the sag and the
tension? Ans. 13.3ft. 90.5 Ibs.
GENERAL PROBLEMS
59
GENERAL PROBLEMS.
Problem 1. Fig. 97 represents a simple triangular truss.* Solve for the
reactions and the stress in each member due to the load.
Ans. Rl = 202 Ibs. R2 = 298 Ibs. AB = 456 Ibs. C. BC = 505 Ibs. C.
AC = 408 Ibs. T.
FIG. 97
Problem 2. Compute the reactions Ri and R2 and the stress in each
member of the truss shown in Fig. 98.
Ans. Ri = R2 = 2500 Ibs. AB = 5000 Ibs. C. AD = 4330 Ibs. T.
BD = 2000 Ibs. T.
Problem 3. The members of the truss shown in Fig. 99 weigh 100 Ibs. per
linear foot. Solve for the reactions and the pin pressure at B in amount and
direction.
Ans. Ri = 929 Ibs. R2 = 871 Ibs. B = 885 Ibs. at 8° 20' with H.
fc
FIG. 99
FIG. 100
Problem 4. Consider the reaction at A of the cantilever truss shown in
Fig. 100 to be horizontal. Solve for the reactions and the internal stresses.
Ans. A = 4000 Ibs. B = 5656 Ibs. at 45° with H. AB = 3000 Ibs. C.
AC = 4472 Ibs. T. BC = 2236 Ibs. C. CD = 2236 Ibs. T. BD =
2000 Ibs. C.
Problem 5. If there is no member AB in the truss of Fig. 100, determine
the amount and direction of the reactions at A and B.
Ans. A = 5000 Ibs. at 36° 50' with H. B = 4120 Ibs. at 14° with H.
Problem 6. In the truss shown in Fig. 101 the reaction of the strut FB
is horizontal. Determine the reactions and the internal stresses.
* Consider all trusses in this set of problems to be pin-connected at all
joints.
60
APPLIED MECHANICS
[CHAP, in
Ans. A = 2655 Ibs. at 57° 20' with H. B = 2433 Ibs. AB = 827 Ibs. T.
AC = 1943 Ibs. T. BC = 1654 Ibs. C. BE = ED = 1000 Ibs. C. CE =
500 Ibs. T. CD = 866 Ibs. T.
R--/0'— -
r-*- 4-M'
FIG. 101
^
FIG. 102
Problem 7. Solve for the reactions and the stress in each member of the
truss shown in Fig. 102.
Ans. & = 11,500 Ibs. R2 = 12,500 Ibs. AB = 13,280 Ibs. C. AC =
6640 Ibs. T. BC = 4042 Ibs. T. BD = 8660 Ibs. C. CD = 2887 Ibs. T.
CE = 7216 Ibs. T. DE = 14,433 Ibs. C.
Problem 8. The roof truss shown in Fig. 103 is held by a hinge at G and is
supported on rollers at A. Solve for the reactions and the internal stresses
caused by the wind loads shown.
Ans. A = 2310 Ibs. Gv = 1155 Ibs. GH = 2000 Ibs. BD = AB = 2888
Ibs. C. AC = CD = 2000 Ibs. T. BC = 2000 Ibs. C. DF = FG = 2310
Ibs. C. CE, DE, EF and EG all equal zero.
— 17s — — TF" rH
£—--/<? --•->)<— /^ '—>|<— I0f— H
FIG. 103
Problem 9. The diagonals BE and CD of the truss shown in Fig. 104 can
take only tensile stress. When the truss is loaded as shown, determine which
diagonal is acting and the amount of the stress in it.
.4ns. CD = 1600 Ibs. T.
Problem 10. A cantilever frame is built up of members pinned together as
shown in Fig. 105. Determine the amount and direction of the reactions at
A and B and the amount of the stress in the diagonals due to the load of 100 Ibs.
at the end.
Ans. A = 608 Ibs. at 9° 25' with H. B = 600 Ibs. hor. CE = 424 Ibs. T.
DE = 424 Ibs. C.
GENERAL PROBLEMS
61
FIG. 105
FIG. 106
Problem 11. Fig. 106 shows a frame supporting a 2-foot pulley. If the
cord CIJ carrying a load of 240 Ibs. is fastened at C, what is the stress in FHt
Am. 170 Ibs. T.
Problem 12. Fig. 107 shows two views of a gin pole
held by three equally spaced guy wires, each at an angle
of 30° with the pole. If the force of the wind is 1000
Ibs. acting at the middle of the pole at right angles
to the vertical plane through guy wire OC, determine
the stress in each wire caused by the wind.
Ans. OA =0. OB = 1154 Ibs. T. OC = 577 Ibs. T.
Problem 13. A bar 10 feet long is held by a pin at
the bottom end and rests at an angle of 45° against a
smooth vertical wall. It carries loads of 100 Ibs., 200
Ibs. and 300 Ibs. at 3 feet, 6 feet and 9 feet respectively
from the lower end. What are the reactions?
Ans. Top, 4201bs.hor. Bottom, 732 Ibs. at 55° with H.
Problem 14. The A-frame shown in Fig. 108 is
pinned at the joints B, C and D, and is supported by
a smooth floor AE. Determine the pin reactions at B,
C and D caused by the 4000 Ibs. load.
FIG.
Ans. B = 1667 Ibs. at 53° 10'
D = 2848 Ibs. at 69° 30' with H.
C
with H. C = 1024 Ibs. at 12° 35' with H.
FIG. 108
Problem 15. Determine the stress in the
brace BE of the crane shown in Fig. 109. De-
termine also the maximum tensile and com- IG*
pressive stresses in the stiff legs DH and DG as the boom rotates about the post.
Ans. BE = 1618 Ibs. DH and DG max. tens, and comp. = 1455 Ibs.
62
APPLIED MECHANICS
[CHAP, in
Problem 16. The simple crane shown in Fig. 110 carries a load of 1000 Ibs.
at the end of the boom. The post AD weighs 600 Ibs., the boom CF 400 Ibs.
and the brace BE 300 Ibs. Determine the reactions at A and D and the pin
pressures at B, C and E.
Ans. A = 2528 Ibs. at 65° 30' with H. D = 1050 Ibs. H. E = 2371 Ibs.
at 42° 35' with H. B = 2584 Ibs. at 47° 20' with H. C = 1761 Ibs. at 6°
30' with H.
Fy*
FIG. 110
FIG. Ill
Problem 17. Determine the stresses in CD and BD, and the reactions at
A and C in the crane shown in Fig. 111. Neglect the weight of the crane
itself.
Ans. CD = 1237 Ibs. T. BD = 1200 Ibs. C. C = 240 Ibs. H. A = 384
Ibs. at 51° 20' with H.
Problem 18. In the crane
shown in Fig. 112, solve for
the stresses in BD and CD
caused by the load of 50,000
Ibs. at D. If the boom
weighs 2000 Ibs. and the
weight on the car is uni-
-6->
I (o>K5)
k- - -so'- —»
FIG. 112
formly distributed, what must the car weigh in order that it does not tip about
point A?
Ans. BD = 100,000 Ibs. C. CD = 86,600 Ibs. T. Wt. = 113,300 Ibs.
D
FIG. 113
Problem 19. Fig. 113 represents a dipper dredge with dimensions as
shown. The boom CG weighs 40,000 Ibs. The handle HF weighs 5000 Ibs.
GENERAL PROBLEMS
63
and in the position shown is at an angle of 15° with the vertical. The dipper
and load weigh 12,000 Ibs. BD is an A-frame, 40 feet in altitude and spread
20 feet at the base. In the filling position consider the pressure to be 10,000
Ibs. at right angles to the handle. Solve for the stresses in the cable FG, the
cables DG and DA, the pin reactions at E and C and the compression in each
member of DB.
Ans. FG = 24,560 Ibs. DG = 41,000 Ibs. DA = 61,700 Ibs. E = 4110
Ibs. at 62° 30' with H. C = 90,100 Ibs. at 39° 40' with H. DB = 23,240
Ibs. in each leg.
SOTons
FIG. 114
Boom
Problem 20. In the derrick shown in Fig. 114, four guy cables, each at an
angle of 30° with the ground, support a mast 55 ft. high. The boom is 85 feet
long. When the boom is in the horizontal position and a load of 50 tons is
being lifted, determine the stresses in the cables FG, cable EG and boom EG.
Determine also the maximum stress that can come upon any one of the four
guy cables as the boom is rotated horizontally.
Ans. FG = 46,000 Ibs. in each cable. Cable EG = 33,330 Ibs.
EG = 187,880 Ibs. C. Max. in guy cable = 178,450 Ibs. T.
Problem 21. The steam hoist represented in Fig. 115 is
raising a weight of 800 Ibs. When the crank is in the position
shown, determine the stress in the connecting rod BD, the
pressure of the guide on the cross-head N and the steam
pressure P for uniform motion.
Ans. BD = 1622 Ibs. N = 270 Ibs. P = 1600 Ibs.
FIG.
FIG.
Problem 22. Determine the stress in the link AB and the shear on the
rivet E of the ice tongs shown in Fig. 116 when supporting a cake of ice weigh-
ing 100 Ibs. Ans. AB = 70.7 Ibs. E = 100 Ibs.
64
APPLIED MECHANICS
[CHAP, in
Problem 23. Determine the stresses in the members AD, DE, BD and BE
of the simple wagon-jack shown in Fig. 117. The jack rests upon the ground
at D and E.
Am. AD = 141 Ibs. T. DE = 100 Ibs. T. BD = 200 Ibs. C. BE = 141
Ibs. C.
FIG. 117
Problem 24. The bridge shown in Fig. 118 has its lower chord in the shape
of a parabola. The total weight of the bridge is 1500 tons. Find the stress
in the lower chord at the piers and at the middle. (The bridge is like the
suspension bridge inverted.)
Ans. 1,352,000 Ibs. at pier. 1,125,000 Ibs. at middle.
_]
FIG. 118
Problem 25. A suspension foot bridge is 80 feet long, 4 feet wide and carries
a load of 100 Ibs. per square foot of floor area. It is supported by two cables
which have 12 ft. of sag. What is the stress in each cable? What is the length
of the cable between supports? Ans. 15,530 Ibs. 84.8 ft.
Problem 26. A wire can safely sustain 70 Ibs. tension. Its weight per
linear foot is 0.025 Ib. If the allowable sag is 1.5 inches, what is the maximum
spacing for posts? If 6 inches sag can be allowed, what is the spacing required?
Ans. 53 ft. 106 ft.
Problem 27. A cable weighing 0.3 Ib. per linear foot is stretched between
posts 160 ft. apart with tension at the middle of 500 Ibs. What is the sag?
What is the amount and direction of the tension at the end?
Ans. 1.92 ft. 500.58 Ibs. 0 = 2° 45' with H.
Problem 28. A cable 800 feet long, weighing 0.5 Ib. per foot, has 250 Ibs.
tension at each end. What is the sag? What is the distance between sup-
ports? Ans. 200 ft. 660 ft. (c = 300 ft.)
CHAPTER IV.
CENTROIDS AND CENTER OF GRAVITY.
43. Centroid of a System of Forces with Fixed Application
Points. In all of the previous discussions of forces applied to
rigid bodies, it has been assumed that the force could be applied at
any point along its line of action. In some cases forces are con-
sidered to be applied at certain definite points which remain fixed,
no matter how the body is displaced or the system of forces
rotated. Consider a system of particles, each of which is acted
upon by a force proportional to its mass, and let these forces be
parallel to each other. It is evident that if the system of par-
ticles is rotated while the forces remain fixed in direction, the
result is the same as if the system of particles remained fixed in
space and the force system were rotated, each force about its
point of application.
Let such a force system be acting upon a system of particles in
the direction of the Y axis. The distance of the resultant from the
XY plane and also from the YZ plane may be determined by the
theorem of moments. Then consider each force of the system to
be rotated about its point of application until the system of forces
is parallel to the X axis. The line of action of the resultant is
necessarily at the same distance from the XY plane that it was
before rotation. Also, its distance from the XZ plane may now be
determined, and its point of intersection with the line of action of
the resultant in its original position must necessarily be the point
about which the resultant was rotated. Next, if from this position
each force is rotated about its point of application until it is
parallel to the Z axis, the line of action of the resultant is neces-
sarily at a fixed distance from the XZ plane during the rotation.
Finally, if from this last position each force is rotated about its
point of application back to its original position parallel to the
Y axis, the line of action of the resultant remains at a fixed dis-
tance from the YZ plane and must necessarily return to its original
position. In order for it to do this, the last two rotations must
necessarily have been made about the same point as the first.
65
66 APPLIED MECHANICS [CHAP, iv
For if the second rotation had been made about a point on the
resultant which had a different X coordinate from the first point
of rotation, the final position of the resultant would have had a
different X coordinate and therefore could not have coincided with
the original position. Similarly if the third rotation had been
made about a point which had a different Z coordinate from the first
point of rotation, the final position of the resultant would have had
a different Z coordinate and therefore could not have coincided
with the original position. This point in the resultant is there-
fore the one fixed point in the system for any possible rotation, and
is called the centroid of the system. Its coordinates are denoted
by x, y, z. (Called "gravity" x, etc.)
Each particle of a body is attracted by the earth, and the force
of this attraction is proportional to the mass of the particle. It is
obvious that the points of application of these forces remain
unchanged for all positions of the body, and that the lines of action
of the forces for bodies of the size considered in engineering prob-
lems are practically parallel. The resultant of all these attractive
forces is called the force of gravity, or the weight of the body, and its
fixed application point is called the center of mass or center of
gravity of the body. Ordinarily it is only necessary to consider
this resultant force.
44. Centroids of Solids, Surfaces and Lines Defined. The
centroid of a geometric solid is that point which coincides with
the center of mass of a homogeneous body occupying the same
volume.
The centroid o!" a surface is the limiting position of the center
of gravity of a homogeneous thin plate, one face of which coincides
with the surface as the thickness of the plate approaches zero.
The centroid of a line is the limiting position of the center of
gravity of a homogeneous thin rod whose axis coincides with the
line as the cross-sectional area of the rod approaches zero.
45. Moment with Respect to a Plane. The moment of a
force with respect to a plane parallel- to its line of action is the
product of the force and the perpendicular distance from the
force to the plane, as discussed in Art. 26. By analogy, the mo-
ment of a solid, surface or line with respect to a plane is equal to the
product of the solid, surface or line and the perpendicular distance
from the plane to its centroid. Since solids, surfaces and lines are
not vector quantities, the sign of the moment must be provided for
ART. 46] CENTROIDS AND CENTER OF GRAVITY 67
by assigning the plus sign to the ordinates on one side of the plane
and the minus sign to those on the other.
By the principle of Art. 26, the moment of the weight of a body
with respect to a plane is equal to the sum of the moments of the weights
of the several particles of the body with respect to the same plane.
For the ZY plane, Wx = Zwx.
For the XY plane, Wz = Smz.
For the XZ plane, Wy = Zwy.
If the moment of the weight of a body with respect to a plane is
zero, the center of gravity of the body is in that plane.
The moment of a solid, surface or line with respect to a plane
is equal to the moment of its separate component parts with
respect to the same plane. For if w is the unit weight of a homo-
geneous body and V is its volume, its total weight is wV = W.
By the above principle,
WVX — WViXi + WV2Xz + WVtfh +, etc.,
or Vx = viXi + v2x2 + ^3X3 +, etc.
Similar propositions hold true for surfaces and lines.
If the moment of a solid, surface or line with respect to a plane
is zero, the centroid is in that plane; and, conversely, if the cen-
troid of a solid, surface or line is in a certain plane of reference,
the moment with respect to that plane is zero.
46. Planes of Symmetry and Axes of Symmetry. If a solid,
surface or line is symmetrical with respect to any plane, the
centroid is in that plane.
If two or more planes of symmetry intersect in a line, this line
is called an axis of symmetry and contains the centroid.
If three or more planes of symmetry intersect each other in a
point, this point is the centroid.
Similar propositions are true for the center of gravity of a mass
if homogeneous.
An observation of the planes of symmetry will enable the cen-
troids of many geometrical figures to be located either partially
or completely. The following are illustrations:
The centroid of a straight line is at its middle point.
The centroid of a circular arc is on the bisecting radius of the arc.
The centroid of a circle or its circumference is the center of the
circle.
The centroid of a parallelogram or its perimeter is the inter-
68 APPLIED MECHANICS [CHAP, iv
section of the lines bisecting the pairs of opposite sides. It is
likewise the intersection of the two diagonals.
The centroid of a sphere or of its surface is the center of the
sphere.
The centroid of a cylinder or of its surface is the middle point of
its axis.
The centroid of a right prism with parallel bases is the middle
point of its axis.
The centroid of a right cone is on its axis.
The centroid of a thin plate is midway between the positions of
the centroids of the faces.
47. Centroid of a System of Forces with Coplanar Applica-
tion Points. In case the application points of the forces of a
system are fixed and coplanar, two moment equations will be
sufficient to locate the centroid if the axes are taken in the plane
of the application points. By Art. 43, the centroid remains fixed
for any rotation of the force system, so the forces may be assumed
to be rotated until they are normal to the plane of the application
points. Then, by the theorem of moments,
XFx
= -~'
SF
The graphic method is readily applied to this case. Assume the
system of forces to be rotated until the forces are in the plane of
the application points, preferably parallel to one of the axes. The
position of the resultant may be determined by the method of
Art. 25. Again assume the system to be rotated through some
angle in its plane, preferably until parallel to the other axis. The
position of the resultant may be determined as before. The
intersection of these two resultants is the position of the centroid.
Problem 1. Parallel forces of 10 Ibs., 24 Ibs., 30 Ibs. and 16 Ibs. have
application points in the XY plane as follows: (2", 3"), (4", 1"), (3", 4") and
(0", 0"). Locate the centroid of the system if all of the forces are in the same
direction. Ans. 2.575", 2.175".
Problem 2. Solve Problem 1 if the first force is reversed in direction.
Ans. 2.767", 1.90".
Problem 3. Solve Problem 1 if both the first and second forces are reversed
in direction. Ans. - 2.17", 5.50".
ART. 48] CENTROIDS AND CENTER OF GRAVITY 69
48. Centroids of Simple Solids and Surfaces. For many
simple surfaces and solids, enough planes or lines containing the
centroid may be determined to locate the centroid completely.
Triangle. The centroid of a triangle is at the intersection of
its medians.
Proof. In Fig. 119, the centroid of any elementary
strip MN parallel to the base and of infinitesimal
width is on the median AD, therefore the centroid
of the triangle is on the median. Likewise it is
on the median BE, and therefore is at their point of
intersection 0.
By geometry, OD = % AD. Therefore the cen-
troid is on any median, at a distance of one-third
its length from its intersection with the base.
The perpendicular distance from 0 to BC is one-
third the altitude of the triangle; therefore, the cen-
troid of a triangle is at the intersection of two lines drawn parallel
respectively to two sides of the triangle and distant one-third of
the altitude from the base.
Slant Area of Pyramid. The centroid of the slant area of a
pyramid is on the axis of the surface, at a distance from the base
equal to one-third of the altitude.
Proof. Consider the pyramid to be cut by planes parallel to
the base and infinitesimal distances apart. The centroid of each
infinitesimal area intercepted between two succeeding planes is on
the axis, therefore the centroid of the total area is on the axis.
The centroid of each of the triangular faces is in a plane distant
one-third of the altitude from the base. Hence the centroid of the
entire slant area is at the intersection of the axis with this plane.
Since the surface of a cone may be considered as the limit of the
surface of a pyramid, the number of whose sides is increased to
infinity, the same proposition holds true for a cone.
Oblique Prism. The centroid of an oblique prism with parallel
bases is at the middle point of its axis.
Proof. Consider the prism to be cut into elementary plates
parallel to the base. The centroid of each plate approaches
coincidence with the centroid of its area as its thickness ap-
proaches zero. The straight line joining these centroids is the
axis of the prism by definition, hence the centroid of the prism is
on its axis. Again, consider the prism to be made up of elementary
70 APPLIED MECHANICS [CHAP.IV
rods parallel to the axis. The centroid of each rod is at its middle
point, hence the centroid of the prism is in the plane passed through
these middle points of the elementary rods parallel to the base.
Oblique Pyramid or Cone. The centroid of an oblique pyramid
or cone is on its axis.
Proof. Consider the pyramid or cone to be cut into elementary
plates parallel to the base. The centroid of each plate approaches
coincidence with the centroid of its area as its thickness ap-
proaches zero. The surface of each plate is an area similar to the
area of the base and its centroid is at a corresponding point in
its area, hence on the axis. Therefore, since the centroids of all
the elementary plates lie upon the axis, the centroid of the entire
pyramid or cone is on the axis.
49. Centroids by Integration. Lines, Plane Surfaces and
Solids. If a solid, surface or line be divided into its infinitesimal
parts, the principle of moments, Art. 26, may be stated as follows :
For solid of volume V,
Vx = xi dVi + xz dV2 + x3 dV3 +, etc., = jxdV;
Vy = fydV-, fv~z = j*z dV. (1)
For surface of area A,
Ax = JxdA; Ay = j ydA'} Az = JzdA. (2)
For line of length I,
Ix = fxdl; Uy = Jydl', Iz = Jzdl. (3)
These expressions may be used when any given solid, surface or
line cannot be divided into finite component parts whose centroids
are known, but is of such form that the differential expression for
the moment can be integrated.
EXAMPLE 1.
Locate the centroid of a circular arc.
X\ Solution: — By symmetry the centroid is on the
x axis OC, Fig. 120, so y = 0. To determine x, use
expression (3).
Ix = fxdl.
I = rot] x = rcos0; dl = rdd.
ART. 49] CENTROIDS AND CENTER OF GRAVITY
71
rax =
+1
r2 cos 0 d0.
rale = r2 sin
rax = 2 r2 sin —
2r . a
x = — sin —
a 2
For a = 360°, sin | = sin 180° = 0.
For a = 180°, sin | = sin 90° = 1.
=0.
x = - = 0.637 r,
For a = 90°, sin- - sin 45° = 0.707. x = 0.707— = 0.901 r.
2 TT
EXAMPLE 2.
Locate the centroid of the sector of a circle.
Solution: — Let the X axis bisect the angle of the sector, Fig. 121. Then
To determine x, use expression (2).
Ax = §xdA.
A = %T*a', dA = pdpdd] x = pcos0.
-JC7
FIG. 121
EXAMPLE 3.
Locate the centroid of the area of a quadrant with
respect to the limiting radius.
Solution: — By Example 2, OA = |£ sin 45°. See Fig.
122.
x = BA = OA sin 45°.
A _ =£r
EXAMPLE 4.
Locate the centroid of a pyramid or cone.
Solution: — By Art. 48 the centroid is on the axis, so it remains to deter-
mine its distance from a plane through the vertex parallel to the base. Let the
FIG. 122
72
APPLIED MECHANICS
[CHAP, iv
pyramid or cone be placed with its vertex at the origin 0, Fig. 123. and its
base MN normal to the X axis. Let A be the area of the base and let a be the
M area of any cross section parallel to the base at
< x .-^JS&rj, distance x from the vertex. Use expression (3).
Vx = fxdV.
-^- x = ( xa dx.
By similar triangles,
6 _ x
B~h
Also by the geometry -of similar areas,
Then
_
A B* h*
A
Ah-
a; = -T A.
4
The distance of the centroid from the base is h.
EXAMPLE 6.
Locate the centroid of a hemisphere.
Solution: — Let the axes be placed as shown in Fig. 124. By sym-
metry y = 0 and z = 0. To determine x, use expression (1).
x = xdV.
V = o71^3! dV = volume of slice AB = -n-y2 dx.
/'
r2 - x2.
3 '
?rS~- T-
3 ~ 2
3
= Cr'xdx- C x3dx.
•/() >/ 0
FIG. 124
Problem 1. Determine by integration the distance of the centroid of an
arc of 90° from the radius at its end. 2 r
Ans. —
7T
ART. 50] CENTROIDS AND CENTER OF GRAVITY 73
Problem 2. Solve Example 2 above by using the elementary sector for dA.
Problem 3. In Example 2 above, use dA = pa dp as
shown by the shaded part in Fig. 125, and solve.
Problem 4. Determine by integration the dis-
tance of the centroid of a quadrant from the limiting
radius.
Problem 5. Locate the centroid of a parabolic seg-
ment of altitude a. The equation of the parabola is
y2 = 4 mx. Ans. x = f a.
Problem 6. Show by integration that the centroid
of a triangle is distant one-third of the altitude from
the base. FlG
50. Centroids of Surfaces and Solids of Revolution. The
centroid of a surface of revolution generated by the rotation of a
line about an axis in its plane is on
the axis. In determining its posi-
tion on the axis, the solution may
be simplified by using for dA the
area generated by the length ds of
the generating line as shown by the-
shaded part in the two views in
Fig. 126.
The centroid of a solid of revolution generated by the rota-
tion of an area about an axis in its plane is on the axis. In
determining its position on the axis, the solution may be sim-
plified by using as dV the volume generated by the element
dA of the generating area.
EXAMPLE 1.
Locate the centroid of a hemispherical surface.
Solution: — Let the axes be placed as shown in Fig. 127. By symmetry
y = 0 and 2=0.
Ax=fxdA. y
A = 2-nr2', dA = 2iryds'} x = rcos0;
y = rsin0; ds = rdB.
FIG. 126
a C2
* = Jo
2 7TT3 cos 0 sin 6 dd.
FIG. 127
74
APPLIED MECHANICS
[CHAP, iv
EXAMPLE 2.
Locate the centroid of a spherical segment.
Solution: — Let the axes be placed as shown in Fig. 128. By symmetry
y = 0 and 2=0.
Vx = fxdV.
dV = inf dx; V = firy* dx; x = r cos0;
dx = —rsinddd; y = rsin0.
_ 3
sin
2 - 3 cos - + cos3 -
FIG. 128
Problem 1. Show by integration that the centroid of the curved surface
of a cone is distant one-third of the altitude from the base.
Problem 2. Locate the centroid of the frustum of a cone which has dimen-
sions as shown in Fig. 129. (Consider the frustum as generated by the
revolution of the shaded trapezoid about the X axis.) Ans. x — 5.9 in.
51. Theorems of Pappus and Guldinus.
I. The area S of the surface generated by
any plane curve revolved about a non-intersect-
ing axis in its plane is equal to the product of
the length of the curve and the length of the path
traced by the centroid of the curve.
Let I be the length of the curve as shown
in Fig. 130.
FIG. 130
fydl
ART. 52] CENT ROWS AND CENTER OF GRAVITY 75
Als°
By eliminating / y dl,
S = 2 iryl
II. The volume V of the solid of revolution generated by a plane
area revolved about a non-intersecting axis in its plane is equal to
the product of the area and the length of
the path traced by the centroid of the
area.
Let A, Fig. 131, be the generating '
area.
r
dA.
Ay = J y
FIG. 131
The volume of the ring generated by the rotation of the differ-
ential area dA is dV and is equal to the product of the length of
the ring and its cross section, 2irydA.
Total volume V = 2w I ydA.
By eliminating I y dA,
V = 2iryA.
Problem 1. Determine the area of a circle by means of Theorem I.
Problem 2. Determine the curved area and the volume of a cone with
height h, radius of base r and slant height I.
Am. Area = irrl. Vol. = ^ irr*h.
o
Problem 3. Determine the curved area and the volume of a cylinder with
height h and radius of base r.
Problem 4. Given the volume of a sphere = ^ irr3, show that the centroid
o
4 r
of a semicircle is 5— from the diameter.
O 7T
52. Center of Gravity of Composite Body. If a body is com-
posed of several simple parts whose centers of gravity are known,
the principle of Art. 26 may be applied.
The moment of a body with respect to a plane is equal to the sum
of the moments of the several parts with respect to the same plane.
Wx = WiXi + w2xz + w&s +, etc.
Similar propositions hold true for y and z. If the body was origi-
nally of simple form with one or more simple parts taken away, a
76
APPLIED MECHANICS
[CHAP, iv
modification of the preceding rule applies. The equation above
may be written
WiXi = Wx — w2X2 — WsXs — , etc.
That' is, the moment of a part of a body with respect to a plane is
equal to the moment of the whole body minus the moment of the parts
taken away.
EXAMPLE 1.
Locate the center of gravity of a wire 12 inches long, bent as shown in
Fig. 132.
Solution: — The weight is proportional to the length.
12z = 3XO + 4X2 + 5X 5.25.
x = 2.85 in.
12jf = 3 X 1.5 + 4 X 0 + 5 X 2.165.
y = 1.28 in.
'— *— x
FIG. 132
FIG. 133
EXAMPLE 2.
Locate the center of gravity of the plate shown in Fig. 133 with respect to
the center of the hole at 0.
Solution: — By symmetry, y = 0. Total original area = 60.56 sq. in.
Area of hole = 3.14 sq. in. Remaining area = 57.42 sq. hi.
57.42 x = 60.56 X 6 - 3.14 X 0.
x = 6.34 in.
Problem 1. Solve for x, y and z of the wire shown in Fig. 132 if the 3-inch
part of the wire is bent forward at right angles to the position shown.
Ans. x = 2.85 in. y = 0.90 in. z = 0.38 in.
Problem 2. Locate the centroid of the area shown in Fig. 134.
Ans. x = 3.58 in. y = -4.07 in.
FIG. 134
FIG. 135
Problem 3. Locate the center of gravity of the gear journal shown in
Fig. 135. Ans. x = 5.425 in.
ART. 54] CENTROIDS AND CENTER OF GRAVITY 77
53. Centroid of Irregular Plane Area. The centroid of any
irregular plane area may be determined graphically. Consider the
standard T-rail section, Fig. 136.
By symmetry, x = 0; y is to be
determined. Only one-half of the
area need be considered, that to
the right of OF. Draw the axes
EF and CD perpendicular to the
Y axis and any distance y\ apart.
Locate as many controlling points
on the bounding line FD as nec-
essary. M and N are two such
points. Determine the points M',
Nf, etc., such that x' = — x. Connect all of the points so obtained
by a smooth curve. Let A be the half area of the rail and A' the
shaded area between the Y axis and the curve M'Nf.
FIG. 136
So
Ay = J ydA.
dA = x dy (horizontal strip).
Ay = I xy dy.
From the relation above, xy = x'yi.
By the substitution of this in the preceding equation,
Ay = 2/1 J x' dy = yiA'.
Then
_ A
Areas should be measured with a planimeter.
Problem 1. Check the graphical method by solving for the location of the
centroid of a rectangle: (1) using h for y\\ (2) using 2 h for yi.
54. Center of Gravity by Experiment. The center of gravity
of an irregular body may be determined by experiment. If the
body is suspended freely, two intersecting vertical planes through
the center of suspension may be marked on the body. Each of
these planes contains the center of gravity, therefore it is in their
line of intersection. If the body is then suspended in some other
position, the intersection of any other vertical plane through the
78
APPLIED MECHANICS
[CHAP, iv
center of support with the line of intersection of the other two
planes determines the center of gravity.
If a body is of such form that it can easily be balanced across a
knife edge, the position of the center of gravity may be determined
readily. The body should be balanced perfectly in some position
and the line of the supporting knife edge marked. The body
should then be rotated and balanced and another line of support
marked. The center of gravity is vertically above the intersection
of the two lines.
GENERAL PROBLEMS.
Problem 1. Locate the centroid of the cross section of the T-bar shown in
Fig. 137. Ans. y = 1.125 in.
{"rod.
(a)
FIG. 137
FIG. 138
Problem 2. Neglecting the fillet and rounded corners, locate the centroid
of the angle section shown in Fig. 138. Ans. x = 2.47 in. y = 1.47 in.
Problem 3. Consider the fillet and rounded corners of the angle section
of Problem 2 and solve for y accurately, (e.g. of shaded area, Fig. 138(b), is
0.22 r from tangent.) Ans. y = 1.46 in.
FIG. 139
Problem 4. Locate the centroid of the channel section shown in Fig. 139.
Ans. y = 0.845 in.
GENERAL PROBLEMS
79
Problem 5. Locate the center of gravity of the governor ball and rod in the
position shown in Fig. 140, The rod is steel and the ball is cast iron.
Ans. x = 10.62 in. y = -10.62 in.
FIG. 140
FIG. 141
Problem 6. Locate the center of gravity of the trapezoidal shaped piece
of sheet iron shown in Fig. 141. Ans. x = 2.44 in. y =2.22 in.
Problem 7. If the triangular ends ABF and CED, Fig. 141, are bent
forward at right angles to the remainder, determine z, y and z.
Ans. x = 2.67 in. y =f 2.22 in. z = 0.37 in.
Problem 8. A controller magnet has dimensions as shown in Fig. 142.
Locate the center of gravity. Ans. y = 2.924 in.
y _.
_L
FIG. 142
Problem 9. An endless wire is bent into the form of an arc of 90° and its
chord. Locate the center of gravity. Ans. 0.808 r from center.
Problem 10. What is the distance from the chord of an arc of 60° to the
centroid of the arc? Ans. 0.0885 r.
Problem 11. A cylinder 6 inches in diameter and 8 inches high has a
cylindrical hole 2 inches in diameter and 4 inches deep bored into its top.
Fig. 143 shows a cross section through the axis of the cylinder and the axis of
the cylindrical hole. The bottom of the hole is conical, each element being at
an angle of 45° with the axis. Locate the center of gravity.
Ans. x = 0.064 in. y = 3.89 in.
Problem 12. A 6-inch cube has cylindrical holes 1 inch in diameter and
2 inches deep drilled in the centers of the top, front and right-hand faces.
Locate the center of gravity. • Ans. x = —0.0149 in.
80
APPLIED MECHANICS
[CHAP, iv
Problem 13. A telephone pole 14 inches in diameter at the bottom and
6 inches at the top is 30 feet long and tapers uniformly. It is loaded upon a
wagon with 4 feet of the butt end projecting in front of the front axle. Where
should the rear axle be placed so as to carry the same weight as the front axle?
Ans. 11.6 ft. from the top end.
Problem 14. A cast iron flywheel 3 feet in diameter has a rim with cross
section as shown in Fig. 144. What is the weight of the rim? Ans. 280 Ibs.
U 4'~.-*\
FIG. 144
FIG. 145
Problem 16. A flywheel 12 feet in diameter has a, rim with cross section as
shown in Fig. 145. What is the volume of the rim? Ans. 19,600 cu. in.
FIG. 146
Problem 16. An idler pulley for a rope drive is 2 feet in diameter and has
a rim with cross section as shown in Fig. 146. The diameter is measured to
the face of the groove. What is the volume of the rim? Ans. 4400 cu. in.
16'xj'pl.
I8"x |V/.
FIG. 147
FIG. 148
Problem 17. Locate the centroid of the cross sectional area of the girder
shown in Fig. 147. The area of one I-beam is 9.26 sq. in.
Ans. y = 7.89 in.
GENERAL PROBLEMS
81
Problem 18. Tne cross section of the end chord of a bridge is shown in
Fig. 148. Determine the position of the centroidal axis parallel to the plate.
The area of one channel is 9.9 sq. in. Ans. y = 10.7 in.
Problem 19. Solve Problem 18, using a 12-in. by 1-in. plate and two
10-in. 20-lb. channels. The area of one channel is 5.88 sq. in.
Ans. y = 7.78 in.
Problem 20.
Fig. 149.
FIG. 149
Locate the centroid of the section of a bulb beam shown in
Ans. x = 2.35 in.
CHAPTER V.
FRICTION.
55. Static and Kinetic Friction. If a block rests upon a
horizontal supporting surface, the weight of the block and the
resistance of the surface are the two forces acting upon the block.
If these distributed forces are considered to be acting at their
centroids, they may be represented by W and N, Fig. 150. If a
small horizontal force P is applied to the block and it is still at rest,
the force to balance P is the resistance of the supporting plane
parallel to P, tangential to the surface, as shown in Fig. 151. This
resistance is called friction and is denoted by F.
wwwpw
\N
I/V
FIG. 150 FIG. 151
If the force P is increased gradually, it will reach a certain value
which the friction F can no longer balance and the block will move.
While the block is at rest the friction is called static friction. The
highest value of the static friction, that when motion is just im-
pending, is called the limiting friction and will be denoted by F'.
After motion begins the friction decreases and is called kinetic
friction, or friction of motion. Friction is always a resisting force
and opposes the motion or the tendency to move.
Adhesion must not be confused with friction. Adhesion is the
attraction between two surfaces in contact. It depends upon the
area in contact and is independent of the pressure. Friction is
independent of the area and varies as the pressure. For nearly
all problems in engineering, adhesion may be neglected.
If the two surfaces in contact are hard and well polished, the
frictional resistance becomes very small but never reaches zero.
For a perfectly smooth surface, then, the resistance would always
82
ART. 56]
FRICTION
83
be normal to the surface. In many problems the friction is very
small compared with the other forces acting and so may be neg-
lected in the solution without appreciable error.
56. Coefficient of Friction. The ratio of the limiting friction
Ff to the normal pressure N is called the coefficient of static friction,
and is denoted by /. In symbols,
-
N
The frictional force F and the normal reaction N acting on the
block in Fig. 152 (a) may be combined into their resultant R. It
is evident that the resultant R must always lean from the normal
in the direction to oppose motion or the tendency to move.
w
-* (a)
FIG. 152
If 0 is the angle between the resultant reaction and the normal,
F
it is plain from Fig. 152 (a) that ~ = tan <£. The maximum value
of 0 corresponding to F' is denoted by <j>' and is called the angle o]
friction. It is evident that / = tan </>'.
If the surface upon which the block rests is inclined at an angle
6 with the horizontal and no force but the pull of gravity and the
reaction of the surface acts upon the block, the angle at which
slipping is impending is 0', the angle of repose. In Fig. 152(b), R
is equal and opposite to W and acts at the angle <£' with the normal,
since slipping impends. From the geometry of the figure, angle
6' = angle <£'.
If the value of the angle $' for two given surfaces is known and
slipping is impending, the resultant reaction becomes known in
direction.
The coefficient of kinetic friction is the ratio of the kinetic friction
F to the normal pressure N and is also denoted by /.
f F
f N'
84
APPLIED MECHANICS
[CHAP, v
EXAMPLE.
A block weighing 500 Ibs. rests upon two wedges which in turn rest upon a
horizontal plane surface. If the angle of the wedges is 10° and the coefficient
of friction is 0.30, what are the forces P, P, required to force the wedges under
the block?
FIG. 153
Graphic Solution: — The angle <£' = tan~* 0.30. = 16° 40'. Fig. 153 (b)
shows the block as a free body. Since slipping is impending, the reactions R\
and ^2 are acting at the angle $' = 16° 40' with the normal to the surface of
contact, or at 26° 40' with the vertical. The force triangle is shown in Fig.
153(c), from which Ri and R2 scale 280 Ibs.
In Fig. 153(d) is shown the left wedge as a free body, with the known force
Ri equal and opposite to Ri acting upon it. The unknown forces are P,
horizontal, and R$ acting to oppose motion at the angle <£' = 16° 40' with the
normal. The force triangle is shown in Fig. 153(e), from which P scales
200 Ibs. and R3 scales 260 Ibs.
Algebraic Solution: — With the 500-lb. weight, Fig. 153(b), as the free body,
equation ^Fx = 0 gives
Rl sin 26° 40' = R2 sin 26° 40'.
Ri = R2.
Equation XFV = 0 gives
2^ cos 26° 40' = 500.
#1 = 280 Ibs.
With the left wedge, Fig. 153(d), as the free body, SFy = 0 gives
280 cos 26° 40' = R3 cos 16° 40'.
Rs = 261 Ibs.
Equation 2FX = 0 gives
P = 280 sin 26° 40' + 261 sin 16° 40'.
P = 126 + 75 = 201 Ibs.
ART. 58] FRICTION 85
Problem 1. In the example above, determine the horizontal force neces-
sary to start the wedge out from under the block.
Ans. 104.1 Ibs. in the reverse direction.
Problem 2. Solve Problem 1 if the coefficient of friction is reduced to 0.15
at all surfaces, Ans. 30.9 Ibs.
67. Laws of Friction. The laws of friction for dry surfaces
were deduced chiefly from the experiments of Morin, Coulomb and
Westinghouse. These may be stated as follows:
1. Friction varies directly as the normal pressure.
2. Limiting static friction is slightly greater than kinetic
friction.
3. Ordinary changes of temperature affect friction only slightly.
4. At slow speeds, friction is independent of the speed. At
high speeds, friction decreases as the speed increases, probably
due to the fact that a film of air is drawn in and acts as a
lubricant.
5. Kinetic friction decreases with the time.
6. Friction is increased by a reversal of motion.
The laws for lubricated surfaces are decidedly different from
those for dry surfaces. For instance, friction is practically in-
dependent of the nature of the surfaces, due to the fact that the
chief friction is between the different layers of the lubricant.
Limiting static friction is much greater than kinetic friction, due
to the fact that while at rest the film of lubricant is pressed out
from between the surfaces. Ordinary changes of temperature
make a decided difference in the character of many lubricants and
therefore affect the amount of friction greatly. Heavy normal
pressure tends to force out the lubricant and therefore increases
the coefficient of friction. As the lubrication becomes poor, the
laws approach those for dry surfaces.
58. Determination of the Coefficient of Friction. The co-
efficient of static friction for two surfaces may be determined
experimentally by finding the pull P necessary to start a weight
W on a horizontal plane, or by finding the angle of inclination of
the plane at which motion is impending for the weight resting
upon it, as explained in Art. 56.
The coefficient of kinetic friction may be determined by finding
the pull P necessary to keep a weight W moving uniformly on a
horizontal plane, or by finding the angle of inclination of the plane
at which the motion of the weight upon it is uniform.
86
APPLIED MECHANICS
[CHAP, v
As would be expected, there are great variations in the values of
the coefficients so obtained. The following table gives the range
of values for the coefficient of static friction for a few materials.
The corresponding coefficients of kinetic friction are 20 per cent
to 40 per cent less than the values for static friction.
SUBSTANCES
STATIC /
Wood on wood
0 30 to 0 70
Metal on metal
0.15 to 0.30
Wood on metal . .
0.20 to 0.60
Leather on wood
0 . 25 to 0 . 50
Leather on metal
0 30 to 0 60
Stone on stone
0 . 40 to 0 . 65
Problem 1. A wooden block weighing 4 Ibs. rests upon a horizontal wooden
table and is just started by a horizontal pull of 2.4 Ibs. Determine /.
Ans. f = 0.6.
Problem 2. A block of cast iron starts to slide upon a steel plate when the
plate is tilted at an angle of 15° with the horizontal. It still slides uniformly
when the plate is lowered to an angle of 11° with the horizontal. What are
the static and kinetic coefficients of friction?
Ans. Static/ = 0.268. Kinetic/ = 0.194.
59. Axle Friction and the Friction Circle. If a cylindrical
axle of radius r rests in a bearing and is rotated, the axle will first
roll from its position of rest until the resultant reaction of the
bearing (resultant of N and F) acts at the angle of friction <£' with
the radius at the point of contact, when slipping of the axle in the
bearing takes place. The circle drawn concentric with the axle
and tangent to the line of this reaction has a radius r sin <f>' and is
called the friction circle.
The radius r and the angle </>' are
usually known, so the friction circle
may be used to locate the point of
contact of the axle and the bearing.
Its chief use is in the graphic solu-
tion. In Fig. 154, Q is the resistance
and P is the working force. These
intersect at B, so the resultant reac-
tion of the bearing must also pass
through B. Since this resultant reaction must also be tangent
to the friction circle, the point of contact of the axle and bearing
is determined.
FIG. 154
ART. 60]
FRICTION
87
To determine the side of the friction circle at which the reaction
is tangent, it is necessary to note the direction of pressure and the
point of contact of the axle with the bearing. The reaction is
tangent to the friction circle on that side toward which the axle
rolls as it rotates.
Another rule is that friction, being a resistance, always shortens
the lever arm of the working force and lengthens the lever arm of
the resisting force.
2'diam.
2 diam.
FIG. 155
Problem 1. Fig. 155 shows a simple steam hoist. Solve for the value of
the force P necessary for uniform motion in the position shown, (1) if friction
is neglected; (2) if friction is considered and/ = 0.15 for all moving surfaces,
Ans. (1) P = 1388 Ibs. (2) P = 1500 Ibs.
Q
W
FIG. 156
Problem 2. Fig. 156 shows the standard compensator for interlocking
signal systems in mean temperature position. Points A and B are fixed. All
pins are 1 inch in diameter. Use / = 0.10 and
determine the value of Q for P = 50 Ibs.
Ans. Q = 49 Ibs.
p
60. Least Pull and Cone of Friction. ^
If the force P, Fig. 157 (a), acts hori-
zontally on a body of weight W, and
motion is impending, the force diagram
is as shown in Fig. 157(b). If W and 0'
are known, R and P can be determined, since for equilibrium the
force polygon must close. If the force P is acting upward at the
(a)
FIG. 157
88
APPLIED MECHANICS
[CHAP, v
W
angle 0 with the horizontal, as in Fig. 158(a), N is decreased and
therefore F' is decreased. Their ratio and the angle <£' remain
constant. From the force diagram,
Fig. 158(b), it is plain that with W
constant and the direction of R con-
stant, the minimum force P to close
the force triangle must be acting at
an angle of 90° with R. So with angle
6 varying the least pull P to start
the block is given when 6 = <j>'.
This result may also be obtained by means of the calculus
method.
If P is acting downward at the angle 6 with the horizontal, N and
Ff are increased, as will be seen in Fig. 159(a) and (b). If the
(a)
FIG. 158
body is free to move in any direction, the cone whose vertex is at
A and whose axis is normal to the surface at A with angle of 2 $'
is called the cone of friction. If the resultant of P and W falls
inside the cone of friction, it is evident that the reaction of the
supporting plane falls within the angle DAE, that is, at an angle 0
with the normal which is less than <£', as shown in Fig. 159(c).
The required frictiorial resistance F is less than the limiting value
F' = N tan <£', hence the plane will hold the body in equilibrium
no matter how much P is increased.
Problem 1. A body weighing 100 Ibs. rests upon a plane surface inclined
at an angle of 10° with the horizontal as shown in Fig. 160.
If / = 0.25, what is the friction under the body? What
force P parallel to the plane will be necessary to start
the body down the plane? What force P parallel to
the plane will be necessary to start the body up the
plane?
FIG. 160
Ans. F = 17 A Ibs. P = 7.2 Ibs. down. P = 42 Ibs. up.
ART. 61]
FRICTION
89
Problem 2. What is the least pull P and its angle with the plane, to
start the body of Problem 1 down the plane? Same for motion up the
plane? Ans. 7 Ibs. at 14° with plane. 40.7 Ibs. at 14° with plane.
Problem 3. In Problem 1, how many degrees each side of the vertical is
the angle for which no motion is possible, no matter how large a downward
force P is applied? Ans. 4° above; 24° below.
61. Rolling Resistance. If the curved surface of a perfect
cylinder touches a perfect plane, they are in contact only along a
line. If a loaded wheel rests upon a rail or roadway, a deformation
is caused so that there is an area of contact. If a horizontal pull
P, Fig. 161, is applied to the axle to move the wheel forward
uniformly, the resultant reaction R of the supporting surface acts
at a point B in front of the vertical radius. Let the horizontal
distance A B be called a. If motion is uniform and if the indenta-
tion is small, equation £ MB = 0 gives, approximately,
Pr = Wa.
If the load W is applied at the circumference of the wheel or
roller, as in Fig. 162, and a force P is applied to move both load
and roller forward uniformly, a similar relation is obtained. Let
a be the distance from the point of application of the resultant to
the vertical radius at the bottom of the roller and ai that at the
top. Then the equation 2M# = 0 gives
If ai = a,
2Pr = W (ai+a).
Wa ,
P = — as before.
r
90 APPLIED MECHANICS [CHAP.V
If the weight W is carried by two or more rollers, R\ + R2 +, etc.
= W (Approx.). Equation %FX = 0 gives
P = (R! + #2+, etc.) sin 0,
if </> is the angle between R and the vertical. Then
P = W sin 0.
If «i = a, P =
2r
Wa
Experiments appear to show that the distance a is practically
constant for the same materials, both for varying loads and varying
radii, within reasonable limits. It is called the coefficient of rolling
friction, or, preferably, the coefficient of rolling resistance. The
experiments of Coulomb, Weisbach and Pambour give the follow-
ing values for a in inches.
WHEEL
TRACK
a (in inches)
Elm
Oak
0 0327
Lignum-VitsB
Oak
0.0195
Cast iron
Cast iron
0.0183
Cast iron or steel
Steel
0 007 to 0 020
Problem 1. If rolling resistance is 1 Ib. per ton for a freight car, what is
the value of the coefficient a for 33-inch wheels? Ans. a = 0.00825 in.
Problem 2. A cast iron engine frame weighing 1200 Ibs. rests upon steel
rollers one inch in diameter which in turn rest upon a pine floor. If the
coefficient of rolling resistance for cast iron upon steel rollers is 0.01 inch, and
that for steel rollers upon pine is 0.03 inch, what horizontal force P is necessary
to move the frame forward uniformly? Ans. P = 48 Ibs.
62. Friction of Brake on Wheel; Graphic Solution. In Fig.
163 (a), the wheel of a car is considered as a free body, with the
external forces acting upon it as shown. In this case the weight
of the wheel is small compared with that of its load, so the force
causing the negative acceleration of the wheel itself is not taken
into account. The wheel is considered to be under static condi-
tions. The normal reaction of the rail is N} equal to the weight
of the wheel and its load; Ni is the normal brake shoe pressure.
Consider the wheel to be rolling, but let the brake shoe pressure be
ART. 62]
FRICTION
91
large enough so that slipping of the wheel is impending. F' is the
limiting static friction between the rail and the wheel, so R acts
at the angle <£' with the normal. As discussed in Art. 56, 0' is
B"
the angle whose tangent is -^ = /. (The coefficient of rolling re-
sistance is comparatively small, and may be neglected.)
FIG. 163
At B, kinetic friction is acting, so F\ = fiNi, and Ri acts at the
angle 0/ with the normal. These two forces meet at C, hence the
reaction of the bearing on the axle must also pass through C. If
the axle friction is neglected, reaction R2 passes through the center
of the axle at 0. If axle friction is considered, the reaction passes
tangent to the friction circle at 0. The graphic solution for all
of the unknown forces is shown in Fig. 163 (b). N, </>' and fa' are
supposed to be known.
The further discussion of the friction of brakes on car wheels and
of friction dynamometers will be given in Chapter XI, Work and
Energy.
FIG. 164
Problem 1. Fig. 164 shows a car on an inclined track, the weight of 400
Ibs. being carried by the two wheels on one side. Consider kinetic / = 0.3
(brake on wheel) and static/ = 0.4 (wheel on rail). Neglect rolling resistance
92
APPLIED MECHANICS
[CHAP, v
and axle friction and determine the two equal and opposite normal brake shoe
pressures, N, N, to allow uniform motion of the car down the plane.
Ans. 173 Ibs.
63. Friction on Pivots. Flat-end Pivot. A flat-end pivot and
its bearing are originally perfect planes, but they cannot remain
so after wear begins. The unit pressure at first is constant over
the whole surface but since the distance traveled over by any
elementary area per revolution varies with its radial distance, the
wear is greater at the outside. This reduces the pressure at the
outside and increases it toward the middle. It is evident that after
the pivot has run until conditions are uniform, the wear parallel
to the axis on the pivot and bearing must be the same at all points.
The wear on any unit area varies both with the distance traveled
(or its radial distance) and with the normal pressure. Therefore,
in order that the wear be uniform over the whole area, it is neces-
sary that the product of the normal pressure on any unit area and
the radial distance of the area shall be constant. If p is the vari-
able unit pressure and p the distance of the unit area from the
center, pp must be constant, or
PP=K.
FIG. 165
FIG. 166
Fig. 165 represents a solid flat-end pivot and Fig. 166 a hollow
flat-end pivot. In either case, dA = p dp dO. The normal pressure
on dA is pp dp dd = K dp dB. The f rictional force on dA is fK dp dd,
f being the coefficient of kinetic friction. The moment of this
frictional force on dA about the center is dM = fKp dp dB. For the
solid pivot of radius r, the total moment about the center is
M = }K
ART. 63] FRICTION 93
As given above, the normal pressure on dA is K dp dO. The total
normal pressure is
P = K f f2"dpd6 = K2irr.
JQ Jo
p
From this. K = - --
Z TTT
By substitution of this value in the expression for the moment
above,
This is seen to be a moment equivalent to the total frictional force
T
fP acting at the mean radius ^ •
z
For the hollow pivot with inner radius r\ and outer radius r2, the
total moment of the frictional force about the center is
M =fK PdpdB = /&(r22 - n2).
*/ri JQ
The normal pressure on dA is K dp dd. The total normal pressure
is
P = K r* f**dpd* = K2<jr(r2 - n).
t/rj JO
P
From this, K = ^-—f - r •
2 7r(r2 - n)
By substitution of this value in the expression for the moment,
As before, this is seen to be a moment equiv-
alent to the total frictional force fP acting at
the mean radius
The collar bearing, shown in Fig. 167, is the ^ J^
same as the hollow pivot. It has the advantage
that it can be placed at any point along the
shaft and also that several can be used on one shaft in order to
obtain any desired amount of bearing area.
Conical Pivot. Fig. 168 represents a conical pivot under axial
load P. Let dP be the load on area dA = p dp dd, and let dN
94
APPLIED MECHANICS
[CHAP, v
be the normal pressure of the bearing on the slant area corre-
sponding. Then since 2Fy = 0,
dNsina = dP,
dP
or
dN =
sin
FIG. 168
The friction caused by the normal pressure
dN isfdN = -. — , and its moment about the
sin a'
center is dM =~ . If p is the variable
sin a
unit pressure on the cross-sectional area, since the same conditions
hold true as in the flat-end pivot,
dP = pp dp dd = K dp de.
The total moment of the frictional forces about the center is
Sin a Sin a Jo «/o
p
As in the flat-end pivot, K = ~ — , so
fPr
sin a
M =
2 sin
Since —
I, the length of an element of the cone of contact, the
sin a
expression for the moment becomes
It will be seen that the moment of the frictional force on a
conical pivot is the same as that on a flat-end pivot whose radius is
equal to the length of the element of the cone of contact.
FIG. 169
64. Friction of Belts. If the belt shown in Fig. 169(a) is
turning the pulley against some resistance, the tension Tz on the
ART. 64] FRICTION 95
driving side is greater than the tension TI on the slack side.
Consider a piece of the belt of ds length as a free body, Fig.
169(b). Let dP be the normal pressure of the pulley on the belt
on ds length. Since the free body is in equilibrium under the
action of the forces shown, equation 2M0 = 0 gives
rdF-rdT = 0.
dF = dT.
By summing forces in the radial direction,
dP = Tsin ^ + (T + dT) sin ^ = - . _ , _ ™ 2
7/j
The term dT sin ^- may be neglected since it is a differential of a
higher order and sin — may be replaced by -~-
Then dP = T d0.
When slipping impends,
dF =fdP.
Therefore dT = f dP = fT dd,
or
By integration, loge ^ = ft.
In terms of common logarithms, this becomes
logiop = 0.4343 ft.
In the exponential form it becomes
s--
The angle 0 is in radians. If the belt is slipping, the same
relations hold true, / being the coefficient of kinetic friction.
These relations are not true if slipping is neither occurring nor
impending.
Problem 1. A belt runs between two pulleys of equal diameter for which
/ = 0.5. If the tension on the slack side of the belt is 100 Ibs., what tension
can be put upon the taut side before slipping is impending? Ans. 481 Ibs.
96 APPLIED MECHANICS [CHAP.V
Problem 2. A windlass has 2£ turns of rope on the drum. / = 0.4 between
the rope and the drum. If the load being pulled is 10,000 Ibs., what tension
must be exerted at the other end to prevent slipping? Ans. 18.7 Ibs.
Problem 3. The tension in the free rope of a block and tackle is 300 Ibs.
It is held by being passed around a post for which/ = 0.3. How many turns
are required to hold it if the tension at the slack end is 5 Ibs.?
Ans. 2.17 turns.
65. Summary of Principles of Friction. In the solution of
problems involving friction, several principles are to be noted
particularly.
(1) If friction is neglected, reactions are always normal to the
surfaces.
(2) If the free body is in motion or tends to move, the friction
of adjoining surfaces upon the free body opposes its motion.
(3) If the free body is at rest and the adjoining surfaces move
or tend to move over it, the friction upon the free body is in the
direction of the moving surface.
(4) The coefficient of static friction is used to determine the
friction only when the body is at rest, with slipping impending.
When slipping is not impending, static conditions determine the
friction.
GENERAL PROBLEMS.
Problem 1. If the static coefficient of friction for cast iron wheels on steel
rails is 0.20, what is the limiting slope down which cars can be run with uni-
form velocity? Ans. 11° 20'.
Problem 2. If the kinetic coefficient of friction for cast iron wheels on
steel rails is 0.15 and the brakes are tightened so that the wheels skid, what is
the unbalanced force down the limiting slope of Problem 1 for a car weighing
100,000 Ibs.? Ans. 4940 Ibs.
Problem 3. Fig. 170 shows a wedge of 20° angle which is forced under a
weight of 200 Ibs. held against a stop block A. If 0' for all surfaces is 15°,
determine the force P necessary to start the wedge under the block.
Ans. 239 Ibs.
GENERAL PROBLEMS
97
Problem 4. Determine the value of the force P of Fig. 170 necessary for
uniform motion of the wedge to the left. Ans. — 35 Ibs.
FIG. 170
FIG. 171
Problem 5. In Fig. 171, A is a 20-lb. body on a 30° plane and B is a 50-lb.
body on a 10° plane. The two are connected by a cord over a pulley at C.
Will the system move if / = I for both bodies? Determine FI, F2 and T.
Ans. F! = 5.77 Ibs. F2 = 12.92 Ibs. T = 4.23 Ibs.
Problem 6. If / = 0.3 for the block shown in Fig. 172 and 0 = 20°, what
is the pressure P necessary to cause motion? Ans. 50 Ibs.
M
10/bs.
T
<- \p
•vjpl ! A
± y
^l-r^
L
N
FIG. 173
FIG. 172
Problem 7. If / = 0.2 for the hanger AB which slides up and down on the
post MN, as shown in Fig. 173, determine how close to the post the load P can
be placed without causing the hanger to slide down. Ans. 3.5 inches.
Problem 8. Use the smallest coefficient of friction for wood on wood as
given in the table and determine the flattest slope a chute for unloading boxes
may have for uniform motion. Ans. 16° 40'.
Problem 9. Determine the constant horizontal force necessary to move a
block of ice weighing 100 Ibs. uniformly up a wooden chute at an angle of 15°
with the horizontal. Use / = 0.05. Ans. 32.2 Ibs.
Problem 10. A plank 12 ft. long rests in a horizontal position upon two
inclined planes, one at 60°, the other at 45° with the horizontal. If <£' = 20°
for each surface, determine the limits of the position where a load can be hung
without causing motion. Neglect the weight of the plank.
Ans. 3.3 ft. from 60° plane. 1 ft. from 45° plane.
Problem 11. If the coefficient of rolling resistance a = 0.01 inch for the
wheels of a freight car on steel rails, and / = 0.03 for axle friction, determine
the horizontal pull necessary to keep a car weighing 100,000 Ibs. in uniform
motion on a level track. The wheels are 33 inches in diameter and the axles
are 4 inches in diameter. Determine also the steepest grade on which the
car would not start. Ans. 424 Ibs. 0.424 per cent.
98 APPLIED MECHANICS [CHAP, v
Problem 12. A white oak beam weighing 200 Ibs. carries a load of 2400
Ibs. and rests on elm rollers 6 inches in diameter. The rollers rest in turn on a
horizontal oak track. What horizontal pull P is necessary to move the beam
and its load? Ans. 28.4 Ibs.
Problem 13. A rectangular block of wood 1 ft. by 1 ft. by 2 ft., weighing 80
Ibs., stands on end on a smooth floor. The coefficient of friction/ = 0.20. If
the block is acted upon by a horizontal force applied at the top of the block till
motion ensues, will the block slide or tip? What is the amount of the force?
Ans. 16 Ibs.
Problem 14. If the same block as in Problem 13 is placed on end on a
plank which is gradually raised at one end, what is the least coefficient of
friction which will make the block tip before sliding? Ans. f = 0.5.
Problem 15. Find the least pull necessary to drag a stone weighing 250 Ibs.
along a horizontal floor for which / = 0.6. Ans. 129 Ibs.
Problem 16. A weight of 10,000 Ibs. is being lowered into the hold of a
vessel. The sustaining rope passes around a spar for which / = 0.2. How
many turns around the spar must the rope have for uniform motion if the
resistance at the other end of the rope is not to exceed 120 Ibs.?
Ans. 3.52 turns.
Problem 17. A rope has 1| turns around the drum of a windlass. If
/ = 0.4 and the pull necessary to keep the rope from slipping is 40 Ibs., what
pull is being exerted at the other end of the rope? Ans. 1730 Ibs.
CHAPTER VI.
MOMENT OF INERTIA,
66. Definition of Moment of Inertia of an Area. Integral
quantities in the form I x2 dA occur in the study of mechanics of
materials. In the expression I x2 dA, dA denotes any differential
area, each part of which is the same distance x from the axis of
reference, called the inertia axis. The sum of these differential
areas equals the total area A. The quantity / x2dA, integrated
between the proper limits, is called the Moment of Inertia* of the
area A.
Defined in words, the Moment of Inertia of a plane area with
respect to any axis is the sum of the products of each elementary
area and the square of its distance from the inertia axis. Moment
of inertia is denoted by I. If it is necessary to specify the axis of
reference (inertia axis), a subscript letter is used, as Ix, IG, etc.
The only axes used are those in the plane of the area and those
* The terms, second moment of area, second moment of mass, etc., would be
preferable, but the term moment of inertia has been in use too long to be
changed. In the case of areas the term is entirely misleading, for since an area
has no inertia it can have no true moment of inertia.
The term was first used by Euler for second moments of mass, on account
of the analogy between rotary and translatory motion.
TTT — — T — Tj-r = acceleration (translatory).
Mass (or Inertia)
Moment of Force
= acceleration (rotary).
J r2 dM (or Moment of Inertia)
According to modern definition, however, inertia is not synonymous with
mass, but is only a property of matter, its amount being proportional to the
mass. For lack of a better name the same term was applied to the expression
x2 dA for areas.
99
100
APPLIED MECHANICS
[CHAP, vi
normal to it. The moment of inertia of an area with respect to an
axis normal to its plane is called the Polar Moment of Inertia.
The expression x2 dA is the product of an area and a distance
squared, hence the moment of inertia of an area is expressed in a
dimension of length raised to the fourth power. In numerical
computations the inch is commonly used as the unit length, and
moment of inertia is, in units of "biquadratic inches," written
in4.
67. Radius of Gyration. It is sometimes convenient to ex-
press a moment of inertia of an area in terms of the area and the
square of a distance. Thus,
/= J*x2dA = k2A.
The quantity "fc" is called the radius of gyration, and is the
distance from the axis at which all the area could be considered as
located and the moment of inertia remain the same. Stated in
another way, k2 is the mean value of x2 for equal differential areas.
As commonly determined,
EXAMPLE 1.
Derive the expressions for the moment of inertia and radius of gyration of
a rectangle which has base b and altitude h, with respect to a centroidal axis
parallel to the base.
Solution: — IY
dA = b dy. See Fig. 174.
The limits of y are — and + - So
b-->
FIG. 174
12
Vl2
ART. 67]
MOMENT OF INERTIA
101
EXAMPLE 2.
Derive the expressions for the moment of inertia and radius of gyration of a
triangle which has base b and altitude h, with respect to its base.
Solution:—
By similar triangles,
= fy*dA.
or
dA = u dy. See Fig. 175.
u h-y
b h '
u-b-r*.
The limits of y are 0 and A. So
r-A A
by*dy- ( j-
•J o n
bh* bh* 1
^"TJ •T~l26fc
A. - \/I - A.
"VA-
FlG- 175
EXAMPLE 3.
Derive the expressions for the moment of inertia and radius of gyration of
a circle of radius r with respect to a diameter.
Solution: — Iv
dA.
dA = Pdp d9. See Fig. 176.
y = p sin 6.
Ix = J* J p3 dp sin2 0 d»,
Cr
J0 P3^P
Polar I0
FIG. 176
EXAMPLE 4.
Derive the expressions for the polar moment of inertia and radius of gyra-
tion of a circle of radius r with respect to an axis through its center.
Solution: — 70 = JV dA.
dA = pdpdd. See Fig. 176.
rr ~zir
70= I I
Jo ^o
= Jirr4 or ^s
102
APPLIED MECHANICS
[CHAP, vi
Problem 1. Derive the expressions for the moment of inertia and radius
of gyration of a rectangle with respect to its base.
Ans. 1=1
3
k = ~
Problem 2. Derive the expressions for the moment of inertia and radius
of gyration of a triangle with respect to its centroidal axis parallel to its base,
Ans. I = bh3.
36
i. _ _ ;
n/ — , —
3V2
Problem 3. Derive the expressions for the moment of inertia and radius
of gyration of a triangle with respect to an axis through the vertex parallel to
the base. Ans. I = \ hh*,
Problem 4. Derive the expressions for the polar moment of inertia and
radius of gyration of a square with respect to an axis through its center.
Ans. I = i 64.
Problem 5. Derive the expressions for the moment of inertia and radius of
gyration of a sector of a circle with respect to its bounding radius.
, _ r*d _ r4 sin 2 0
= 8 16
68. Sign of Moment of Inertia. Since the squares of both
positive and negative quantities are positive and since areas are
always positive, it follows that moments of inertia are always
positive. In Fig. 177 (a), Ix of the area M for which all of the
values of y are positive is the same as Ix of the symmetrical area
N for which all of the values of y are negative. Thus the moment
of inertia of each half of a circular area is the same with respect to
any diameter.
FIG. 177
The value of y does not depend upon the position of the area
along the X axis. For example, Ix may be computed separately
for the two triangles P and Q, Fig. 177(b), or the triangles may be
considered to be shifted until they touch and form the one triangle
P'Q' for which Ix may be computed with one operation.
69. Relation Between Moments of Inertia with Respect to
Two Parallel Axes in the Plane of the Area. If the moment of
ART. 691 MOMENT OF INERTIA 103
inertia of an area with respect to an axis other than a centroidal
axis is required, it may be obtained without integration if the
moment of inertia with respect to the parallel centroidal axis is
known. In Fig. 178 let X0 be a centroidal axis and Xi any other
axis parallel to it, in the plane of the area, at a distance d from the
centroidal axis.
IXi = CifdA+2d Cy dA + d2 CdA.
lXl = Iz. + Ad2. FIG. 178
2 d I y dA = 0 because I y dA = yA, and for the centroidal axis
y = o.
Stated in words the equation above is as follows. The moment
of inertia of an area with respect to any axis in its plane is equal to its
moment of inertia with respect to a parallel centroidal axis plus the
product of the area and the square of the distance between the axes.
This equation is commonly known as the Transfer Formula.
If both sides of the equation IXl = /z, + Ad2 are divided by A,
it becomes
III - lls + d2
A ~~ A
Hence kx2 = kXn2 + d2}
or k2 = k2 + d2.
Problem 1. Given !XQ for a rectangle = fa bh3, derive the expression for
7 with respect to the base. Ans. I = % bh3.
Problem 2. Determine the moment of inertia of a rectangle 20 in. by 1 in.
with respect to an axis in its plane parallel to the 20-inch side and 10 inches
from the centroidal axis. Ans. 7 = 2001.67 in.4.
Problem 3. Given 7 = \ bh3 for a triangle with respect to an axis through
the vertex parallel to the base, derive the expression for 7 (1) with respect to
the base; (2) with respect to a centroidal axis parallel to the base.
Ans. (1) 7 = & bh\ (2) 7 = ^ ^3-
Problem 4. Given 7 of a semicircle with respect to its bounding diameter,
derive the expression for 7 Yo with respect to a parallel centroidal axis.
Problem 5. Derive the expression for 7 of a circle with respect to a tangent.
Ans. 7 = £ Trr4.
104
APPLIED MECHANICS
[CHAP, vi
70. Relation Between Moments of Inertia with Respect to
Three Rectangular Axes. Let Fig. 179 represent any plane area
and let Z be the polar axis through 0.
Then
-Y So
FIG. 179
Iz=
r* = x2 + y2.
Iz = f(x2 + 2/2) dA
= Cx2dA + Cy*dA.
The polar moment of inertia of an area with respect to any axis
equals the sum of its moments of inertia with respect to any two rec-
tangular axes in the area intersecting the polar axis.
Problem 1. Show that for a circle which has its center at the origin,
'*-*/*.
Problem 2. Derive the expression for the polar moment of inertia of a
square with respect to an axis through one corner.
Problem 3. Prove that the moment of inertia of a square with respect to
any centroidal axis in its plane is a constant.
Problem 4. Derive the expression for the moment of inertia of a circle
with respect to a polar axis intersecting its circumference.
71. Relation Between Polar Moments of Inertia with Respect
to Parallel Axes. The relation between the polar moment of
inertia of an area with respect to a centroidal axis and that with
respect to any parallel axis is similar to that between moments of
inertia with respect to parallel axes in the plane of the area.
Let XQ and F0, Fig. 180, be the centroidal
axes and X and Y any other parallel axes, all
of them being in the plane of the area. By
Art. 69,
y
and
I = 7
Fo
FIG. 180
Let Z be the axis through 0 perpendicular to X and Y and let
ZQ be the axis through C perpendicular to X0 and Y0. By Art. 70,
ART. 72]
Since
MOMENT OF INERTIA
and
105
The polar moment of inertia of an area with respect to any axis is
equal to its polar moment of inertia with respect to the centroidal axis
plus the product of the area and the square of the distance between the
two axes.
Problem 1. Solve Problem 4 of Art. 70 by the method of this Article.
Problem 2. Derive the expression for the polar moment of inertia of a
square with respect to an axis bisecting a side. Ans. I = •£% b*.
72. Moment of Inertia of Composite Areas. The moment of
inertia of a composite area with respect to any axis equals the sum
of the moments of inertia of the separate parts with respect to the
same axis. For example, the moment of inertia of the trapezoid
ABCD, Fig. 181, with respect to the base AD is the sum of the
FIG. 182
moments of inertia of the rectangle FBCE and the two triangles
ABF and ECD with respect to AD. The moment of inertia of the
annulus, Fig. 182, with respect to a diameter is equal to the moment
of inertia of the larger circle minus the moment of inertia of the
smaller circle with respect to the same axis.
Ix = I7rr24 - iTrri4.
Problem 1. In Fig. 181, let BC = 4 in., BF = 5 in., AF = 3 in. and
ED = 2 in. Determine the moment of inertia of the trapezoid with respect
to the base AD and transfer to its parallel centroidal axis GG.
Ans. IG = 64.25 in.4, (y = 2.18 in.)
Problem 2. Solve for IG of the trapezoid of Problem 1 by getting first the
moment of inertia of each of the component parts with respect to its own
centroidal axis parallel to the axis GG and then transferring to axis GG.
Problem 3. In Fig. 181 consider a semicircle of 2 inches radius with center
at F cut out of the original trapezoid. Determine Ix and Iy of the remaining
area.
Ans.
Ix = 212.47 in.4.
IY = 221.64 in.4.
106
APPLIED MECHANICS
[CHAP, vi
Problem 4. A wooden box girder is made of four 2 in. by 10 in. planks as
shown hi Fig. 183. Compute the moment of inertia of the cross section with
respect to axis 1-1. Ans. Ii-i = 1786.67 in.4.
Problem 5. Compute the moment of inertia of the cross section of half-
pipe shown in Fig. 184, with respect to its centroidal axis GG parallel to the
bounding diameter. Ans. 1 = 7210 in.4, (y = 14.67 in.)
W-.l
k 10--
FIG. 183
73. Moment of Inertia with Respect to Inclined Axes. In
Fig. 185, let X and Y be any two rectangular axes for which
Ix = I y2 dA and IY = / x2 dA. X' and Y' are axes at the angle s
6 with the original pair. Then
2 dA and
Also y' = y cos 0 — x sin 0 and x' = x cos 0 + y sin 0, from the
geometry of the figure. By squaring these values and substituting
above,
Ix, = ly2cos28dA-2 I xycosdsiuddA -f lx2sm2ddA.
By integration,
Ix> = cos2e-Ix + sin20-7y -2cos0sin0 I xy dA.
1 + cos20 1 -cos20
cos20 =
and
2 cos 0 sin 0 = sin 2 0.
By substitution of these values the equation above becomes
+
+ -^~2 — -cos20 - sin 2 0 / xy dA. (1)
ART. 74] MOMENT OF INERTIA 107
Similarly,
7 Ix + IY
ly — 7^
cos20 + sin20 I xy dA. (2)
By adding these expressions for Ixf and IY' it is found that
Ix' + IY' = Ix + IY,
as shown in Art. 70.
Equations (1) and (2) simplify what would otherwise be a very
tedious operation. If the moments of inertia of an area with
respect to any two rectangular axes in the plane of the area are
known, the moment of inertia with respect to any coplanar in-
clined axis passing through their point of intersection may be
easily computed.
EXAMPLE.
Determine the moment of inertia of a rectangle 6 inches wide and 2 inches
high with respect to an axis through the lower left-hand corner at an angle of
15° with the base, as in Fig. 186.
Solution: — Ix = 16. Iy = 144.
//~G j~2
xydA= I I xdxydy = 36.
»/ 0 »/ 0
Jx' = 80-64 cos 30° - 36 sin 30°.
Ix' = 6.58 in.4. FIG. 186
Problem 1. Solve for IY' of the rectangle of Fig. 186. Ans. 153.4 in.4.
Problem 2. Determine Ix' of a 3-inch square with respect to a diagonal.
Ans. 6.75. in.4.
Problem 3. Solve for Ix' of the rectangle of Fig. 186 if 0 = — 15°.
Ans. 42.6 in.4.
Problem 4. A rectangle 2 inches wide and 4 inches high is placed with the
origin of coordinates 2 inches below the lower left-hand corner. Determine
Ix' for the axis through this origin at an angle of 15° with the horizontal.
Ans. 114.1 in.4.
74. Product of Inertia. By analogy with moment of inertia,
the expression I xy dA is called the Product of Inertia of the area
and is denoted by H. The form of the expression shows that
product of inertia is always taken with respect to a pair of rec-
tangular axes.
// either one of the axes is an axis of symmetry for the area, the
product of inertia with respect to that pair of axes is zero.
108
APPLIED MECHANICS
[CHAP, vi
Proof: — Let Fig. 187 be any area symmetrical with respect to
the Y axis.
H = fxydA.
H
0.
In the summation of the products xy dA it
will be seen that for each term (-\-x)ydA
there is a numerically equal term (— x) y dA
to neutralize it. Hence for a figure sym-
x metrical with respect to the Y axis,
I xydA
Similarly, H = 0 for a figure symmetrical with respect to the X
axis.
75. Relation Between Products of Inertia with Respect to
Parallel Axes. After the product of inertia
is determined with respect to a pair of
rectangular centroidal axes, it may be calcu-
lated easily with respect to any other pair of
parallel axes.
In Fig. 188, OX and OY are any two
rectangular centroidal axes, O'X' and O'Y*
are any other pair of parallel axes in the same
plane, (x, y) are the coordinates of dA with
respect to the original axes and (m + x, n + y)
the coordinates of dA with respect to the new axes.
+ x) (n + y) dA.
Ho = I mndA + I mydA + I nxdA + / xydA.
Ho' = mnA + 0 + 0 + H0.
HO is the product of inertia of the area with respect to the
original axes.
This expression is similar to the transfer formula for moment of
inertia, d2 being replaced by mn.
The quantities m and n may be either positive or negative, so the
term mnA may be either positive or negative. If the centroid of
the area is in the first or third quadrant of the axes with respect to
which H is taken, mnA is positive; if in the second or fourth
quadrant it4 is negative.
FIG. 188
ART. 75]
MOMENT OF INERTIA
109
As in the case of moment of inertia, the
product of inertia of an area composed of
several simple parts with respect to any
pair of axes is equal to the algebraic sum
of the products of inertia of the several
parts with respect to the same axes.
For example, if HXY of the angle section
of Fig. 189 is required, the area may be
divided into the two rectangles M and N.
Then HXY of the angle section = HXY
M
N
\
I
FIG. 189
of M + HXY of N.
EXAMPLE.
Determine the value of H0' for the right triangle shown in Fig. 190.
H0 = fxydA
= 4.5 in.4.
Ho' = Ho + mnA
= +4.5 + 36
= +40.5 in.4.
In this case H0' can be determined more easily by integrating directly than
by the transfer method used above.
#„'= C C xdxydy
Jo JQ
= +40.5 in.4.
If Ho = 0, as it does if either X or Y is an axis of symmetry, the transfer
method is much simpler, for then
HQ = mnA
k
Fia. 190
FIG. 191
Problem 1. Determine the product of inertia of the 6 in. by 1 in. rectangle
shown in Fig. 191 with respect to the X and Y axes. Ans. +9 in.4.
110 APPLIED MECHANICS [CHAP, vi
Problem 2. Locate the centroidal axes parallel to the legs of the
6" X 6" XI" angle section, Fig. 189, and calculate the product of inertia
of the section with respect to these axes. Ans. +20.44 in.4.
Problem 3. Determine the product of inertia of a 4" X 3" X \" angle
section with respect to the centroidal axes parallel to the legs.
Ans. ±2.019 in 4.
76. Maximum and Minimum Moments of Inertia. The
moment of inertia of an area with respect to an axis at an angle 6
with some original axis is given by Equation (1), Art. 73.
6.
As 6 varies the value of Ix varies. The values of 6 for maximum
and minimum values of Ix are determined by differentiating the
expression for Ixf and placing the first derivative equal to zero.
~£ = (7y - Ix) sin 20 - 2#cos20.
OAJ
For the maximum or minimum value of Ix ', —jjr = 0. Then
au
Two values of 2 6 differing by 180° are obtained from the equa-
tion above, and therefore two values of 6 differing by 90°. One
value gives the angle for maximum Ix'j the other the value for
minimum Ix'. The maximum and minimum moments of inertia
are called the principal moments of inertia, and the corresponding
axes the principal axes.
If either the X or Y axis is an axis of symmetry, H = 0, by Art.
74, therefore tan 2 6 = 0. 2 6 = 0° or 180° and 0 = 0° or 90°, so
the X and Y axes are the principal axes.
Problem 1. Determine the maximum and minimum moments of inertia
of the rectangle shown in Fig. 191 with respect to axes through the lower left-
hand corner. Ans. Max. / = 73. 14 in.4. Mm. / = 0.86 in.4.
Problem 2. Determine the maximum and minimum moments of inertia of
a 4" X 3" X \" angle section with respect to centroidal axes.
Ans. Max. / = 6.14 in.4. Min. / = 1.33 in.4.
Problem 3. Determine the maximum and minimum moments of inertia of
a 6" X 6" X \" angle section with respect to centroidal axes.
Ans. Max. / = 31.75 in.4. Min. / = 8.07 in.4.
ART. 77]
MOMENT OF INERTIA
111
77. Moment of Inertia of Mass. The moment of inertia of a
body with respect to any axis is the sum of the products of each
elementary mass and the square of its distance from the axis.
The same notation is used for moment of inertia of masses as
was used for moment of inertia of areas, with the addition of M
for mass, V for volume and 7 for mass per unit volume.
M =
=fx*dM = y Cx*dV.
Units of Moment of Inertia of Masses. The moment of inertia of
a body is in terms of a length squared and a mass. Since the unit
of mass commonly used in engineering is one containing g units of
weight and g is usually given in units of feet per second per second,
all dimensions should be in feet. No name has been given to the
unit moment of inertia of mass.
EXAMPLE 1.
Show that for a right prism of altitude h, with respect to an axis perpendicu-
lar to the base,
I = yh X Polar I of Base.
Solution: — In Fig. 192, let the Y axis be the inertia
axis of the prism. The mass of the elementary prism
whose altitude is h and base dA is dM — yhdA.
IY = yh X Polar I of Base.
Right Circular Cylinder. Since the polar moment
of inertia of a circle with respect to its center is ^ irr4,
the moment of inertia of a cylinder of radius r and
altitude h with respect to its geometric axis is given by
k== \^- = ~
"•M -s/2
EXAMPLE 2.
Show that for a homogeneous sphere of radius r, with respect to a diameter,
/ = •= Mr2 and k = r
o
N/l
Solution: — In Fig. 193 let the Y axis be the inertia axis. Let the sphere
112
APPLIED MECHANICS
[CHAP, vi
be divided into thin plates by planes perpendicular to the F axis, each of
thickness dy and of radius r\. One plate is shown at A.
n2 = r2 - y\
and dM = yirr^ dy — yir (r2 — y2) dy.
By Example 1 the moment of inertia of this thin plate with respect to the
Y axis is
Y =
J i
If these differential moments of inertia are summed between the limits — r
and +r, the entire moment of inertia of the sphere is obtained.
2 4
7 = ^ Mr2, since M = yV = 7 ~ 7rr3>
u o
FIG. 193
EXAMPLE 3.
Derive the expression for the moment of inertia of a circular plate of radius
r and thickness dt with respect to a centroidal diameter.
Solution: — Fig. 194(a) is a top view and Fig. 194(b) is an edge view of the
plate. Let each dM be a prism of volume p dp dd dt whose distance from the X
axis is y.
dM = 7 dV = yp dp dd dt; y = p sin 6.
J
sin2 Byp dp dd dt,
/»V /»r
/ = ydt\ I sin20d0P3dp
A Jo -A)
Mr2.
ART. 78]
MOMENT OF INERTIA
113
EXAMPLE 4.
Derive the expression for the moment of inertia of a slender rod with respect
to an axis through one end.
Solution: — Let L be the length of the rod, Fig. 195, W its weight, M its
mass, w its weight per linear unit and 0 its angle with the Y axis.
y
Then
By integration,
r2 = Z2 sin2 6. dM = y dW = 'yw dl.
Jy = Cp gin2 6yw dl = yw sin2 0 J* /2 dl.
I = yw sin2 0 -o- = Q •W£2 sin2 0.
o o
If the axis is normal to the rod, 0 = 90° and 7 = - ML2.
o
FIG. 195
Problem 1. Derive the expressions for the moment of inertia and radius of
gyration of a homogeneous parallelepiped whose sides are a, b and c, with
respect to a geometric axis parallel to side c. . _ M . 2
Problem 2. Derive the expressions for the moment of inertia and radius of
gyration of a homogeneous right circular cone with respect to its geometric
axis. The radius of the base is r and the altitude is h. Ans. I = T3^ Mr2.
Problem 3. Determine the moment of inertia of a cast iron cylinder 2 feet
in diameter and 6 inches high with respect to its geometric axis.
Ans. I = 10.97.
Problem 4. Determine the moment of inertia of a cast iron governor ball
4 inches in diameter with respect to its diameter. Ans. I = 0.00301.
78. Relation Between Moments of Inertia of Mass with
Respect to Parallel Axes. The moment of inertia of a body with
respect to any axis is equal to the moment of inertia with respect to a
parallel centroidal axis plus the product of the
mass of the body and the square of the distance
between the axes.
In symbols,
I = IG + Md2.
Proof: — Fig. 196 represents a section of the
body perpendicular to the inertia axis which
FIG. 196 passes through G. Let 0 be the point where
any parallel axis cuts the section. Then with respect to the axis
through 0,
/ = Co2dM.
114 APPLIED MECHANICS
From the figure,
[CHAP, vi
Then
/ = f (a* + yt) dM + f (a2 + 62) dM + |*2 az dM + f 2 by dM
= IG + Md2 + 0 + 0. (See Arts. 69 and 70.)
By dividing by M,
or
X1
EXAMPLE.
Show that for a right circular cylinder with radius r and altitude h, the
moment of inertia with respect to a centroidal axis
parallel to the base is
(r2 /,2 >
1 + k
Solution: — Consider the cylinder to be divided into
circular plates, each of thickness dy, one of which is
shown at A in Fig. 197. The moment of inertia of
plate A with respect to its own central axis X' is \ dMr2,
as shown in Art. 77, and with respect to axis X is $
dMr2 + dMy2. dM = yirr2 dy, so the moment of inertia
of all the plates, or the entire cylinder, with respect to
axis X is
.h h
FIG. 197
1 C
A fyjrr4 1
4 J
C
1
»/
By integration,
ix-
Problem 1. Derive the expression for the moment of inertia of a right
circular cone of height h and radius of base r with respect to an axis through
the vertex parallel to the base. T _ 3 , , /r2
- 5^4 +
Problem 2. A rectangular parallelepiped has sides a, b and c. Derive the
expression for its moment of inertia with respect .to the central axis in the ac
face, parallel to c. T ,, /a2 .
Ans. I = M ( ^2 +
Problem 3. Derive the expression for the moment of inertia of a sphere
with respect to a tangent.
Ans. I = ^ Mr2.
ART. 79] MOMENT OF INERTIA 115
Problem 4. Derive the expressions for the moment of inertia and radius of
gyration of a right circular cylinder with respect to a diameter of the base.
Ans. 7 =
Note: — If r2 is negligible compared with h2, the expression for / becomes
f Mb?, as in Example 4, Art. 77.
Problem 5. Determine the moment of inertia of a steel cylinder 2 inches in
diameter and 4 feet long with respect to a diameter of one end.
Ans. / = 7.08.
Problem 6. Determine the moment of inertia of a 3-inch cast iron governor
ball with respect to an axis 8 inches from its center. Ans. I = 0.0515.
79. Determination of Moment of Inertia by Experiment. If
the form of a body is such that its moment of inertia cannot be
computed readily by integration, it may be determined experi-
mentally with a fair degree of accuracy. There are several
methods but the one most readily applicable to problems occurring
in engineering is the pendulum method. As will be shown in Art.
108, the radius of gyration of a compound pendulum is given by
in which T is the time of one complete oscillation, g is the accelera-
tion of gravity and d is the distance from the axis of rotation to
the parallel centroidal axis. The axis of rotation must be parallel
to the axis for which the moment of inertia is required. If the
body is vibrated and time T of one oscillation determined, k may
be computed. Then 70 = Mk2, in which 70 is the moment of
inertia with respect to the axis of rotation. The moment of
inertia with respect to the parallel centroidal axis is given by
IG = Jo - Md\
From this, if desired, the moment of inertia with respect to any
parallel axis may be computed.
Problem 1. A pair of 33-inch cast iron freight car wheels and their con-
necting axle weighed 700 Ibs. When suspended from knife edges 4 feet from
the axis of the wheels they vibrated 100 times (complete oscillations) in 3
minutes and 43.7 seconds. Determine / and k with respect to their centroidal
axis. Ans. 7 = 6.99. k = 0.568 ft.
Problem 2. The connecting rod of a Corliss engine weighed 267 Ibs. Its
center of gravity was 48.5 inches from the crosshead pin. When suspended
from the crosshead end it vibrated 40 times (complete oscillations) in 96 seconds.
Determine 7 with respect to the axis of the crosshead pin.
Ans. 7 = 157.4.
116
APPLIED MECHANICS
[CHAP, vi
GENERAL PROBLEMS.
Structural Steel Shapes and Built-up Sections. .
Problem 1. Locate the position of the gravity axes parallel to the legs of
a 6" X 3£" X f" angle section and compute the moment of inertia with
respect to each axis. Ans. 7,_i = 3.342 in.4. /2_2 = 12.864 in.4.
Note: — Fillets and rounded corners are neglected.
Problem 2. The dimensions of a standard 12-in. 3L5-lb. I-beam are given
in Fig. 198. Compute the moment of inertia and radius of gyration with
respect to the centroidal axes 1-1 and 2-2.
Ans. A-! = 216.15 in.4, /a-a = 9.51 in.4.
.
2— 10
FIG, 199
Problem 3. Structural steel Handbooks give the following formula for the
moment of inertia of an I-beam with respect to axis 1-1, Fig. 199.
Derive this formula.
Problem 4. Compute the moment of inertia of the symmetrical Z-bar
shown in Fig. 200 with respect to axes 1-1 and 2-2.
Ans. 7i-i = 42.12 in.4. /.._2 = 15.44 in.4.
FIG. 200
FIG. 201
Problem 5. The built-up I-section shown in Fig. 201 consists of four
6" X 3£" X I" flange angles and one 12 in. by I in. web plate. Height h' =
12£ in. Compute the moment of inertia with respect to axes 1-1 and 2-2.
Ans. /i-i = 456.7 in.4. 72_2 = 119.3 in.4.
MOMENT OF INERTIA
117
Problem 6. If a 14 in. by $• in. flange plate is added to each flange of the I-
section of Problem 5, what is the moment of inertia of the section with respect
to axis 1-1? Ans. 7i_i = 1026.0 in.4.
Problem 7. In Problem 5, what difference is made if an 18 in. by f in
web plate is substituted for the 12 in. by f in. plate? Height h' = 18£ in.
Ans. li-i is 2 3 times as much. 72-2 is changed only in the second
decimal place.
•Z
Given, IH ofll-bearit,=?/5.8in*
FIG. 202
Given I, .of I Channel =402.7 in.4
I2.2ofl Channel ll.??m.4
Area of I Channel = 14. 71 sq. in.
FIG. 203
Problem 8. Two standard 12-in. 31.5-lb. I-beams are used with two 14 in.
by | in. plates to form a box-girder, Fig. 202. Compute the moment of iner-
tia of the section with respect to axis 1-1. Ans. li-i = 1286.6 in.4.
Problem 9. A column is formed by two 15-in. 50-lb. channels and two
20 in. by \ in. plates, as shown in Fig. 203. Compute the moment of inertia
of the section with respect to each axis.
Ans. A-! = 2007.1 in.4. /a-2 = 2074.6 in.4.
Problem 10.
in., flange angles 6" X 6" X *
Area- 8.44
sq.in.
FIG. 204
The plate girder shown in Fig. 204 has a web plate 48 in. by |
*" and flange plates 14 in. by 1 in. Compute /i-i.
Ans. /i_! = 37,406.6 in.4.
118
APPLIED MECHANICS
[CHAP, vi
Moment of Inertia of Masses.
Problem 11. A steel disk 30 inches in diameter and 3 inches thick has a
cylindrical hole 4 inches in diameter at the center and another 2 inches in
diameter 12 inches from the center. What is the moment of inertia with
respect to the geometric axis? Ans. 14.5016.
' —
•>
h
^
i
I
L
0
f
c4>
«i
T
'B
FIG. 205
re'L>\
A
l^x
>"V-.
A
!
1
1
f-
5s
rA'.
A
C\j
IV
N
^ CVj
</ \\s\\ ^
i
V
;v
•V
-v
3-
.
r
L
->
?c
1"
b
r
"1
._,
— .
\\
L
FIG. 206
FIG. 207
Problem 12. A flywheel governor consists of a cast iron plate 4 in. by
1 in. by 22 in., Fig. 205, which connects the cast iron cylinders A and B.
The cylinder at A is 8 inches in diameter and 2 inches thick. The one at B is
8 inches in diameter and 4 inches thick. Compute the moment of inertia of
the governor with respect to an axis through 0 parallel to the axes of the
cylinders. Ans. 70 = 4.797.
Problem 13. A cast iron pulley with a solid web has dimensions as shown
in Fig. 206. Compute its moment of inertia with respect to the axis of rota-
tion. Ans. 7 = 4.531.
Problem 14. Compute the moment of inertia of the cast iron flywheel
shown in Fig. 207. The wheel has six elliptical spokes which may be con-
sidered as slender rods. Ans. / = 74.67.
PART II. KINETICS.
CHAPTER VII.
RECTILINEAR MOTION.
80. Velocity and Speed. The velocity of a particle is its rate
of motion with respect to an assumed point of reference. (Any
body which is small compared to its range of motion is considered
to be a particle.) The point of reference usually assumed is some
point which is at rest with respect to the surface of the earth.
Motion is said to be rectilinear if the path of the particle is a straight
line, and curvilinear if the path is a curved line. Curvilinear
motion will be discussed in Chapter VIII.
If a particle in rectilinear motion traverses equal spaces in equal
time intervals, its velocity is uniform and is equal to the ratio of
any given space s to the time t in which it was traversed, or
If a particle moves over unequal spaces in equal time intervals, its
velocity is variable. In this case the ratio of any given space s to
the time t in which it was traversed gives only the average velocity.
As the space s is shortened until it becomes ds, this average velocity
approaches the value of the instantaneous velocity at the point
where ds is taken. This instantaneous velocity is
ds
V=dt'
Velocity has direction as well as magnitude. It is therefore a
vector quantity and is represented graphically by a vector. Speed
is the scalar or quantity part of velocity and is merely the rate of
travel, irrespective of direction.
The units of velocity and speed are any units of length and time,
as feet per second, miles per hour, etc.
119
120 APPLIED MECHANICS [CHAP, vn
Problem 1. Reduce velocity of 60 miles per hour to terms of feet per second.
Ans. 88 ft. per sec.
Problem 2. A man runs 100 yards in 10 seconds. What is his average
speed in miles per hour? Ans. 20.45 mi. per hr.
Problem 3. If in a certain motion of a body, s = 4 ts, s being in feet and t
in seconds, what is the velocity of the body at the end of 4 seconds?
Ans. v4 = 192 ft. per sec.
81. Acceleration. A cceleration is the rate of change of velocity.
If the velocity is constant, the acceleration is of course zero. If
the velocity is changed by equal amounts in equal time intervals,
the acceleration is constant; if by unequal amounts in equal time
intervals, it is variable. When the velocity increases, the accelera-
tion is usually called positive; when the velocity decreases the
acceleration is usually called negative.
If the acceleration is constant, it is the total change in velocity
during unit time and its amount is obtained by dividing the total
change in velocity by the time t in which the change was made.
If a represents the acceleration, v0 the initial velocity and v the
final velocity,
a ^ V- Vo
If Vo = 0,
_ y
If the acceleration is variable, its value at any instant is given
by the ratio of the infinitesimal change in velocity dv to the corre-
sponding time dt, or
dv d2s
~ dt ~ dt2'
By eliminating dt between the equations v = -r. and a = -r,
ctt dt
there is obtained the important relation,
v dv = a ds.
The units used are those of velocity and time. If the velocity
of a body changes from 0 to 20 ft. per second in 4 seconds, its
acceleration is a velocity change of 5 ft. per second in a second, or
as commonly written, 5 ft. per sec. per sec. If the velocity of a
body decreases from 40 miles per hour to 20 miles per hour in 10
seconds, the acceleration is —2 miles per hour per second.
ART. 82] RECTILINEAR MOTION 121
Acceleration, like velocity, is a vector quantity and is repre-
sented graphically by a vector.
Problem 1. A street car attains a velocity of 10 miles per hour in 4 seconds.
What is its average acceleration in feet per second per second?
Ans. 3f ft. per sec. per sec.
Problem 2. If the piston of a steam engine attains a velocity of 3 feet per
second in ^ of a second, what is its average acceleration? What is the space
traversed if the acceleration is constant?
Ans. a = 120 ft. per sec. per sec. s = 0.45 inch.
Problem 3. If in a certain motion of a body s = 4 tz, s being in feet and t
in seconds, what is the acceleration of the body at the end of 4 seconds?
Ans. a4 = 96 ft. per sec. per sec.
82. Constant Acceleration. The integration of the differen-
tial expression a = -r between the proper limits gives the velocity
in terms of the time. A second integration gives the distance in
terms of the time.
dv , J±
a = -T., or dv = a dt.
at
Let VQ be the initial velocity. Then, if a is constant,
rdv = a I dt',
*^o
v — VQ = at',
or v = v0 + at. (1)
ds
Since v — -^ , ds = VQ dt + at dt.
at
Ids = v0 I dt + a I tdt.
t/o t/o t/o
ro t/o
s = v0t + \ at2. (2)
The expression v dv = ads may be integrated in a similar manner.
/ v dv = a I dsj
Jv t/0
V2
or v2 = v02 + 2 as. (3)
If flo = 0, v2 = 2 as.
Equations (1), (2) and (3) may be derived by simple algebra, as
follows. If a particle gains a velocity of a during each unit of
122 APPLIED MECHANICS [CHAP, vn
time, in t units of time it will have gained at units of velocity. Its
final velocity, then, will be the sum of its initial velocity v0, and
the velocity it has gained, at.
v = v0 + at. (1)
The average velocity during time t will be the mean of its initial
velocity VQ and its final velocity v0 + at. The average velocity is
— 2 — = VQ -\- % at. The space passed over will be given by the
product of the average velocity and the time, or
g = V+JJo t = Vot + i ^ (2)
If t is eliminated from equations (1) and (2), the resulting equation
is v2 = v02 + 2 as. (3)
Problem 1. A car traveling at the rate of 40 feet per second is brought to
rest in a distance of 100 feet. What is the average acceleration?
Ans. a = —8 ft. per sec. per sec.
Problem 2. A ball has an initial velocity of 10 feet per second and an
acceleration of 4 feet per second per second. What is the velocity at the end
of 4 seconds? At the end of 5 seconds? What is the space passed over during
the 5th second? Ans. v4 = 26 ft. per sec. v5 = 30 ft. per sec. s = 28 ft.
83. Falling Bodies, Air Neglected. For comparatively short
falls of bodies near the surface of the earth, the attraction of the
earth may be considered constant; consequently, if the resistance
of the air is neglected, the acceleration caused by this attraction
may also be considered constant. This acceleration, denoted by
g, is approximately 32.2 ft. per sec. per sec., and if no other value
is given, this should be used in the solution of all problems. (The
accurate value is given by
g = 32.0894 (1 + 0.0052375 sin2 1) (1 - 0.0000000957 h),
in which I is the latitude in degrees and h is the elevation above
sea level, in feet.)
The equations of motion derived in the preceding article become,
for falling bodies,
V = V0 + gt. (1)
h = v0t + i gt2. (2)
v2 = vo2 + 2 gh. (3)
ART. 83] RECTILINEAR MOTION 123
Space, velocity and acceleration are all positive downward. If
the body falls from rest, VQ = 0, hence v = V2 gh. If the body is
projected vertically upward, VQ is negative. The body rises t — -
y
seconds through a distance h = — ^- where it comes to rest. It
then falls from rest and passes the initial point with a velocity of
-4-Vo and continues downward from that point exactly as though
projected downward with the same velocity.
EXAMPLE.
A ball is shot upward with a velocity of 30 ft. per sec. One second later
another is shot upward with a velocity of 100 ft. per sec. Where and when do
they pass?
Solution: — Let Si be the distance from the starting point to the point where
they pass, t\ the time elapsing after the first is discharged, and tz the time
elapsing after the second is discharged.
Then
b = ti- 1.
The initial velocity v0 is negative in each case. After ti seconds the first ball
will be a distance from the starting point, s = wt + \ gt2, or Si = — 30 ti +
16.1 if.
At the same instant the second ball, which has been traveling only ti — 1
seconds, will be a distance from the starting point, s = vrf + I fl^2, or
s2 = -100ft- 1) + 16.1 (<! - I)2.
When the two balls pass, si = s2, so
-30 1, + 16.1 h2 = -100 (ti - 1) + 16.1 (ti - I)2.
ti = 1.137 sec.
si = s2 = -13.3ft.
It will be noticed that the first ball has reached the top point in its path in
30
75777, or 0.932 sec., and is therefore moving downward when they pass.
6Z.Z
Problem 1. If a body falls freely from rest, what is its velocity 5 seconds
later? How far has it fallen? Ans. v = 161 ft. per sec. s = 402.5 ft.
Problem 2. A body falls from a table to the floor, a distance of 3 feet.
What is the velocity with which it strikes the floor? What is the time re-
quired? Ans. v = 13.9 ft. per sec. t = 0.431 sec.
Problem 3. The striking velocity of a pile driver hammer is to be 50 feet
per second. From what height must it be dropped? Ans. 38.8 ft.
Problem 4. From a mine cage which is descending with a velocity of 20
feet per second, a body rolls off and falls 200 feet to the bottom of the shaft.
With what velocity does it strike and what is the time required?
Ans. v = 115.2 ft. per sec. t = 2.96 sec.
Problem 6. In Problem 4, substitute " ascending " for " descending " and
solve. Ans. v = 115.2 it. per sec. t = 4.20 sec.
124 APPLIED MECHANICS [CHAP, vn
84. Relation Between Force, Mass and Acceleration. New-
ton's Second Law of Motion, Art. 6, states that the accelerations
of bodies are directly proportional to the resultant forces acting
and inversely proportional to the masses acted upon. Let F be
the resultant force which acts upon mass M to produce accelera-
F
tion a. Then a varies as — , or F varies as Ma.
m'
F = KMa,
K being a constant, the value of which depends upon the units
used. In American engineering practice the unit of force used is
the pound and the unit of acceleration is the foot per second per
second. In order to make the constant K = 1 and thus simplify
the expression, the unit of mass used is that amount in which unit
force produces unit acceleration. If a resultant force of one pound
acts upon a quantity of matter weighing one pound, the accelera-
tion produced is 32.2 feet per second per second as in the case
of a falling body. If the quantity of matter is increased and the
force remains constant, the acceleration will decrease proportion-
ately, so if the quantity of matter weighs 32.2 pounds, the force of
one pound will produce an acceleration of one foot per second per
second. It is seen, then, that 32.2 pounds of matter is the unit
of mass in which unit force produces unit acceleration and if this
unit is used, K — 1. In order to obtain the number of mass units
in a given quantity of matter, its weight in pounds must be divided
W
by 32.2. That is, M = — • Then
w
F = Ma = —a.
g
In the equation F = Ma, M is a scalar quantity and F and a
are vector quantities. If F\ and F2 are any two components into
which the force F may be resolved, and a\ and «2 are the corre-
sponding components of the acceleration, parallel respectively to
FI and Fzt it follows that
Fi = Ma} and F2 = Ma*.
If the components are the rectangular components Fx and Fy,
Fx = Max and Fy = May.
ART. 85] RECTILINEAR MOTION 125
EXAMPLE.
A horizontal force of 10 pounds is exerted upon a body whose weight is
100 pounds and which is resting upon a smooth horizontal surface. What is
the velocity of the body at the end of 5 seconds, and what is the distance
passed over?
W
Solution: — F = Ma = — a.
a = 3.22 ft. per sec. per sec.
The force is constant, so the body has uniformly accelerated motion. Since
the body starts from rest, the equations of motion are
v = at and s = % at2.
v = 3.22 X 5 = 16.1 ft. per sec.
s = | X3.22 X25 = 40.25 ft.
Problem 1. A resultant force of 50 pounds acts for 4 seconds upon a body
weighing 200 pounds. What is the acceleration of the body and the space
passed over during that time?
Ans. a = 8.05 ft. per sec. per sec. s = 64.4 ft.
Problem 2. An elevator which weighs 1000 pounds starts from rest and in
2 seconds has attained a velocity of 10 feet per second upward, with uniform
acceleration. What is the tension T in the supporting cables?
Ans. Total T = 1155 Ibs.
Problem 3. If in Problem 2 the tension is reduced to 900 pounds, in what
time will the elevator come to rest? Ans. 3.1 sec.
Problem 4. A body weighing 10 pounds is projected down a smooth 45°
plane with an initial velocity of 3 feet per second. How far will it move
during the third second? If at the end of the third second a constant resisting
force of 15 pounds begins to act, how long will it move before coming to rest?
Ans. s3 = 59.91 ft. 2.8 sec.
85. Effective Forces; D'Alembert's Principle. In general, any
particle of a body considered free has a system of forces acting upon
it, some of which may be external to the body as a whole and some
of which are internal. The resultant of all these forces for the
particle is called the effective force for the particle, and is equal to
dM • a, dM being the mass of the particle and a its acceleration. If
the particles of the body were all made free of each other and each
had its effective force acting, the motion of the system of particles
would be the same as the actual motion of the body. The result-
ant of all these effective forces for all the particles of the body is
called the resultant effective force for the body.
126 APPLIED MECHANICS [CHAP. VH
Since the internal forces between the particles of a rigid body are
always mutual, that is, equal and opposite, their total resultant for
the whole body is zero. It follows then that the resultant effective
force for all the particles of a rigid body must be equivalent to the
resultant of the external forces. If F is the resultant of the external
forces,
F = fdMa.
If the motion is translation, a is the same in amount and direc-
tion for all of the particles, so, for translation, F = aU dM = Ma.
Since, each particle has a force equivalent to dM • a acting upon it
and since each force is proportional to the mass of the particle, the
point of application of the resultant is necessarily the same as that
of a system of particles acted upon by their own weights. As
shown in Art. 43, this is the mass center of the body.
If the system of effective forces were reversed and added to the
external system of forces, the result would be equilibrium without
changing any of the external forces.
This principle is called D'Alembert's Principle and is applicable
to both rigid and non-rigid bodies, but only in the case of rigid
bodies is it sufficient to determine the motion.
By this method a problem in kinetics is reduced to a simpler one
in statics, for then all the equations of static equilibrium will apply;
I>FX = 0, SFV = 0 and 2 Mom. = 0. The student should keep in
mind that this is only an imaginary force system, added to the
actual system for the purposes of solution.
(a) (b)
FIG. 208
This method of procedure does not conflict with the method of
Art. 84, as will now be shown. In Fig. 208 (a), let F be the result-
ant force acting upon mass M. From Art. 84, the force F pro-
duces an acceleration a in the mass M of such an amount that
F = Ma = — a.
g
In Fig. 208 (b), F is the resultant force acting upon mass M. The
ART. 85]
RECTILINEAR MOTION
127
W
resultant effective force — a is reversed and added to the system to
y
produce a condition of equilibrium. Since the system of forces
acting upon the mass M is now in equilibrium, %FX = 0, so
or
as before. The reversed effective force is sometimes called the
inertia force of the body, since it may be considered as a force
resisting the change in velocity.
In a problem of this kind, no advantage is gained by the use of
this method. If the solution of a problem requires the use of a
moment equation, however, the addition of the reversed effective
force simplifies the solution very much, as will be shown in the
following Example.
EXAMPLE.
A safe with weight and dimensions as shown in Fig. 209(a) is pulled along
a horizontal track by a force of 100 pounds. A force of 60 pounds is sufficient
to move it uniformly. Determine the normal components of the reactions at
A and B.
££
(a)
FIG. 209
Solution: — Fig. 209 (b) shows the free body diagram, with all the external
W
forces acting and in addition the reversed effective force — a, acting through the
y
center of gravity. The free body now has a balanced system of forces acting
and the equations of equilibrium are true.
Since 60 pounds will move the body uniformly, Fi + F2 = 60.
W
" x~~g°"
100 - 60 = — a = 40.
9
128 APPLIED MECHANICS [CHAP, vn
The equation 2MA = 0 gives
100 X 1 - 40 X 3 - 1000 X 1 + #2 X 2 = 0.
R2 = 510 Ibs.
From the equation 2FV = 0,
#1 = 490 Ibs.
Problem 1. In the example above, consider the 100-pound force to be
removed while the resisting frictional forces Fi and F2 remain the same.
Determine Ri and Rz while the safe is coming to rest.
Ans. Ri = 590 Ibs. #2 = 410 Ibs.
Problem 2. A 50-ton car moving with a speed of 30 miles per hour is
brought to rest by means of the brakes in 6 seconds. The height of the center
of gravity of the car above the track is 5 feet and the distance between the
trucks is 30 feet. Find the pressure on each truck.
Ans. 53,793 Ibs. on front. 46,207 Ibs. on rear.
Problem 3. A rectangular block 1 foot square and 4 feet long stands on
end on the flat surface of a car with its sides parallel to the motion of the
car. What acceleration may be given to the car before the block tips if the
friction is sufficient to prevent sliding?
Ans. Limiting a = 8.05 ft. per sec. per sec.
86. Composition and Resolution of Velocities and Accelera-
tions. As stated previously, velocity and acceleration are vector
quantities and can therefore be compounded
or resolved the same as forces and displace-
ments. (See Chapter I.) If a particle at point
FIG 210 ^' ^g< 210' *s Displaced first to B, then to D,
its resultant displacement is AD. If the dis-
placements were in the other order, the path would be A CD and
the displacement would be AD as before. If the displacements
occurred simultaneously, the path would be AD.
If AB represents a force acting upon a body at A, and AC repre-
sents another acting upon the same body, the resultant force, if
they are simultaneous, is represented by AD.
The forces represented by AB and AC produce corresponding
proportional accelerations, which to some scale are also represented
by vectors AB and AC, and if the two are simultaneous, the result-
ant acceleration is given by the vector AD.
If the forces act during time t, each produces its corresponding
velocity, v = at. If not simultaneous, the vectors AB and AC
represent to some scale the velocities acquired. If they are
simultaneous, the vector AD represents the resultant velocity.
ART. 87] RECTILINEAR MOTION 129
In the discussion of some problems it is often much simpler to
consider the velocity or acceleration as made up of certain com-
ponents. Thus a velocity is often resolved into its x and y com-
ponents, an acceleration into its normal and tangential components,
etc.
Problem 1. A body has a velocity of 120 feet per second directed at an
angle of 45° with the horizontal, as shown in Fig. 211. Resolve the velocity
into horizontal and vertical components. Resolve
into components parallel and perpendicular respec-
tively to the 30° line, AB.
Ans. Vx = vv = 84.84 ft. per sec. 115.90 ft. per
sec. 31.06 ft. per sec.
Problem 2. A body has an acceleration verti-
cally downward of 32.2 feet per second per second -pIG
and an acceleration horizontally of 4 feet per second
per second. What is the resultant acceleration of the body? What is the
velocity of the body at the end of 0.3 of a second?
Ans. a = 32.45 ft. per sec. per sec. at 7° 05' with vertical, v = 9.74 ft. per
sec. at 7° 05' with vertical.
87. Acceleration Varying with Distance; Simple Harmonic
Motion. If the acceleration of a particle is variable, it may be
given in terms of the velocity, of the time, or of the distance from
some fixed point. In any case, it is necessary to analyze the
motion by means of the equations,
ds dv d2s
" = It> n = di = dp and "*'-«*•
If a particle has a motion along a straight line with an accelera-
tion always directed toward a fixed point on the line and propor-
tional to its displacement from that point in either direction, its
motion is vibratory and is called Simple Harmonic Motion. Since
the acceleration is always toward a fixed point in its path, it is
necessarily directed oppositely to its displacement; that is, if the
displacement is to the right or positive, the
acceleration is to the left or negative.
M 0 A N *n -^S- 212> let the particle be at A with
FIG 212 displacement s from the fixed point 0, and let
it be moving to the right with velocity v and
acceleration —a. The limiting values of s are +s' and — s', and
the velocity of the particle at 0 is v0.
From the definition,
a = -Ks,
130 APPLIED MECHANICS [CHAP, vn
K being a constant representing the acceleration at unit distance
from 0. Then
v dv = a ds = — Ks ds.
£vdv = —K I sds.
Jo
& - v<? = -Ks2.
v = Vv02 - Ks2. (1)
ds ds / — rr~:
v==di> so Jt = vv° Ks'
r* c*
dt =
Jo Jo
ds
° VvQ2-Ks2
= visi -v-
Solving for s,
= VQ .
" VKSI
From equation (1). s' = ± --S=, since v = 0 when s = s'. The
VK
time for the particle to move from 0 to N is obtained from equa-
tion (2) by letting s = s'.
•i
IN~VKSI ~2\/F
The time for the particle to move from 0 to N and back to 0 again
is obtained from equation (2) by placing s = 0.
1 TT
to — —7-= sin"1 0 = — 7='
VK VK
It is seen from the two expressions above that the time required
for the particle to move from 0 to N is the same as that required
for it to move from N to 0. Also, motion to the left of 0 corre-
sponds exactly to that to the right of 0, so the time of one complete
vibration is given by
The period is independent of the displacement. Equation (1)
gives the velocity in terms of the distance and equation (2) gives
ART. 87] RECTILINEAR MOTION 131
the time in terms of the distance. By eliminating s between these
two equations, the relation between velocity and time is obtained.
V = VQ COS t Vlt. (5)
It will be seen from equation (5) that VQ is the maximum value of v.
Simple harmonic motion may be illustrated by means of a ball
placed between two horizontal springs and supported by a smooth
plane, as shown in Fig. 213. The ball is attached to both springs
so that when it is displaced, one spring is compressed and the
other is elongated. At the middle position 0, neither spring is
acting. If displaced a distance sf and then released, it will vibrate
back and forth from -f-s' to — sf. This motion would continue
indefinitely if there were no friction of the supporting surface, the
air and the springs.
EXAMPLE.
Let each spring in Fig. 213 be a 20-lb. spring and let the ball weigh 10 pounds.
(A 20-lb. spring is one which is compressed or elongated 1 inch by a static load
of 20 pounds.) If the ball is displaced s' = +2 inches
and then released, what is the value of y0? What is
the period of vibration T? Find the position, velocity
and acceleration of the ball 0.25 of a second after
release. FIG. 213
Solution: — The constant K is the value of the
acceleration when s = 1 ft. The force exerted by each spring is 240 times the
displacement s in feet, so the total force exerted by the two springs would be
480 pounds when s = 1 foot.
F = — a, so 01 = K = ^ = 1545.6.
From equation (4), period T = —== — 0.16 sec.
VK
Since v = 0 when s = s', equation (1) gives
v0 = VRs' = 39.3 X I = 6.55 ft. per sec.
In 0.25 of a second the ball has made one complete vibration, with 0.09 of a
second remaining. Time is measured from the position 0 when the ball is
moving to the right, so t = 0.09 + 0.04 = 0.13 sec.
= sin (0.13 X 39.3)
= -0.1537ft.
132 APPLIED MECHANICS [CHAP.VII
From equation (5) the velocity at the end of 0.25 of a second after release is
v = v0 cos 292°.8.
v = 6.55 X 0.388 = 2.54 ft. per sec.
The acceleration at the end of 0.25 of a second after release is given by the
equation
a = -Ks = -1545.6 X (-0.1537) = 237 ft. per sec. per sec.
Problem 1. If in the apparatus of Fig. 213 the ball is not attached to the
springs, what will be its period of vibration? If s' = +3 inches, find v0 and
the position, velocity and acceleration \ second after release.
Arts. T = 0.226 sec. v0 = 6.95 ft. per sec. s = +0.0588 ft. v = -6.75ft.
per sec. a = —45.4 ft. per sec. per sec.
88. Acceleration Varying with Distance : Direct Solution. If
a body weighing W pounds rests upon a coil spring of scale Q, the
W
spring will be compressed a distance d = 0 n feet. This position
iZ (^
is called the static position. If the body falls through a distance
h and strikes the coil spring, the motion below the static position is
simple harmonic motion, and if the body should become attached
to the spring the total ensuing motion would be simple harmonic.
In a case of this kind, however, it is simpler to integrate directly
for the equations of motion, and use the initial position of the
upper end of the unstressed spring as the origin.
EXAMPLE.
A ball weighing 100 pounds falls through a distance of 2 feet and strikes
upon a 1000 Ib. spring. See Fig. 214. Determine the compression of the
spring, the total time to rest, the maximum velocity and the
maximum acceleration.
Solution: — Since the scale of the spring is 1000 Ibs. per
inch, the resistance of the spring against the ball is 12,000 s, s
being in feet. The total force acting upon the ball after it
strikes the spring is 100 - 12,000 s. From F = — a,
g — 120 gs.
F10- 214 vdv = ads=gds — 120 gs ds.
Let the velocity at the instant of striking be VL
f v dv = g f ds - 120 g f s ds.
J V\ •'0 "'O
(1)
ART. 89] RECTILINEAR MOTION 133
\
From the laws of falling bodies, Vi = V2 gh = 11.35 ft. per sec. At maximum
compression, v = 0, so s' = 0.1911 ft. From equation (1),
ds
Since v = -T. »
at
v = V + 2 gs — 120 0s2.
Cdt = C ds -•
Jo Jo vV + 2gs — 120 gs'-
,--L— in-. 240^-20 -.«
• V1200 V402 + 480s»1Uo
Fors=s'- ^
From the equation s = | <#2, the time <' to fall 2 feet = 0.352 sec., so the total
time from release till it comes to rest at the bottom of its travel is t + t' =
0.026 + 0.352 = 0.378 sec.
In order to obtain the value of s for which the velocity is a maximum. -^
as
from equation (1) is equated to zero. For maximum velocity, s = T|7 ft.
Maximum v = 028.8 + 2~^JP - \MX120 = 1L36 ft> per Sec" The
maximum acceleration is at the bottom of its travel. From the equation for
a given above, maximum a = 32.2 - 120 X 32.2 X 0.191 = -706.2 ft. per
sec. per sec.
Problem 1. Solve the Example above by the equations of harmonic motion.
Problem 2. A 100,000-lb. freight car, equipped with an 80,000-lb. spring,
strikes a bumping post while moving with a velocity of 2 miles per hour.
Find the compression of the spring, assuming that it alone is deformed. (The
equations derived above will not apply to this case but similar equations must
be integrated.) Am. 2 inches.
89. Motion in which Acceleration Varies
Inversely as the Square of the Distance.
) If a body is at a distance s from the cen-
ter of the earth, s being aggreciabfa larger
than r, the radius of the earth, the iefce of at-
traction of the earth varies inversely as the
square of the distance. Let Fig. 215 represent
a diametral section of the earth with a body
at point A, distant s from the center, and let
a be its acceleration. Let the center of the
earth be used as the origin and let distance, velocity and accelera-
tion be considered positive outward.
134 APPLIED MECHANICS [CHAP, vn
Then - = — . Since the acceleration is negative toward the
9 s
earth,
|M
vdv = ads — —g — ds.
s
rCT
vdv = -gr2 I s~2ds.
*/«
If s = oo , v = V2~gr, so if a body fell toward^ the earth from an
infinite distance, its velocity would be \/2 gr = 6.95 miles per
second for r = 3960 miles. If falling from any finite distance s,
the velocity must be less than this. If projected outward with
this velocity, the body would be carried to an infinite distance.
Problem 1. Find the velocity acquired by a body in falling 1000 miles to
the surface of the earth. Use r = 4000 miles. Ans. 3.12 miles per sec.
90. Relative Motion. By velocity of a body is usually meant
the velocity of the body with respect to the point on the earth from
which the motion is observed. Although any point on the earth
has several motions in space, it is considered to be at rest and the
motion of any body relative to that point on the earth is called its
absolute velocity. The velocity of one body with respect to another
body is called its relative velocity.
Let A, Fig. 216, represent a car, top view, and B a body on the
fi R car. If the car A moves into the posi-
, ~| tion A' in one second, BB2 is the
"c-klr I amount of its velocity and also its dis-
placement. If the body B moves from
the position B to B± relative to the car
while the car has moved from A to A', the vector BBf, the result-
ant of BBi and BB2, gives the absolute velocity and displacement
of B.
Similarly, if vector BB2 represents to some scale the absolute
acceleration of the car A, and vector BBi represents the relative
ART. 90J RECTILINEAR MOTION 135
acceleration of B with respect to A, the vector BE' represents the
absolute acceleration of B.
If either the absolute velocity or acceleration of A or the relative
velocity or acceleration of B with respect to A is a variable, the
absolute velocity or acceleration of B is given by the vector sum
of the corresponding instantaneous values of the components.
This principle may be formulated as follows: The absolute
displacement, velocity or acceleration of any body plus (vectori-
ally) the relative displacement, velocity or acceleration of another
body with respect to the first, equals the absolute displacement,
velocity or acceleration respectively of the second. Stated more
briefly,
The Absolute of A + the Relative of B to A = the Absolute of B.
If any two of these three quantities are known, the other may be
found.
EXAMPLE 1.
A man swims across a stream which flows at the rate of 2 ft. per second.
If he can swim at the rate of 3 feet per second, in what direction must he swim
in order to land directly opposite? If the stream is 1000
feet wide, find the time to cross.
Solution: — If 0, Fig. 217, is the point from which the
swimmer starts, OC is the required direction of his absolute
velocity. OA, 2 units to scale, represents the absolute
velocity of the stream. Then OB, 3 units to scale, the
vector representing the relative velocity of the swimmer
with respect to the stream, must be at such an angle that
their resultant lies along OC. The graphical construc-
tion gives angle COB = 42°, or by trigonometry, sin COB must equal i so
COB = 41° 49'.
Length OC is the absolute distance traveled by the swimmer in one second.
OC = VI = 2.236 feet. The time to cross is t = ^^ = 447 sec. =7 min.
A.ZiJ<d
27 sec.
EXAMPLE 2.
An ice boat runs due east with a velocity of 30 miles per hour. The wind
blows from the northwest with a velocity of 20 miles per hour. How can the
sail be set so that a forward pressure will be exerted?
Solution: — In this problem the two absolute velocities are given, to find
the velocity of the wind relative to the boat. From 0, Fig. 218, lay down the
vectors representing the absolute velocities of the boat and the wind. Join
the ends of the vectors with the line AB, and through 0 draw tha vector OC,
equal and parallel to AB. The vector OC represents to scale the velocity of
the wind relative to the boat. Angle COD = 41° 40'. If now the sail is set
136
APPLIED MECHANICS
[CHAP, vii
in some such position as MN, at an angle with the axis of the boat less than
41° 40', the wind striking it in the direction OC will exert a small forward
thrust. If this thrust is equal to the frictional resistance, the velocity will be
maintained, while if it is greater, the velocity will be still further increased.
It is thus seen that an ice boat may travel faster than the wind which drives it,
a fact which has often been proved experimentally.
FIG. 219
Problem 1. Water enters an inward flow radial impulse turbine, Fig. 219.
at an angle of 45° with the radius produced. If its velocity is 100 feet per
second and the rim velocity of the wheel is 60 feet per second, what should be
the angle of the outer edge of the vane in order that the water may enter
smoothly? What is the initial relative velocity?
Ans. 8° 35' with the radius. 71.5 ft. per sec.
FIG. 220
Problem 2. In Fig. 220, AB is the connecting rod and BO the crank of a
reciprocating engine. If the velocity of the crank pin B is 10 feet per second,
what is the velocity of the crosshead A when the crank is at an angle of 45°
with the horizontal? (The relative velocity of B with respect to A must
necessarily be normal to AB, since AB is a rigid body.)
Ans. 8.36 ft. per sec.
RECTILINEAR MOTION 137
GENERAL PROBLEMS.
Problem 1. In an elevator shaft 200 feet high an elevator is moving
upward with a velocity of 10 feet per second. At the instant it is 6 feet from
the bottom a ball drops from the top of the shaft. When and where will the
ball and the elevator meet and with what relative velocity?
Ans. 3.17 sec. 37.74 ft. from bottom. Rel. v = 112.2 ft. per sec.
Problem 2. From the top of a tower 120 feet high a ball is dropped at the
same instant that another is shot upward from the bottom with a velocity of
100 feet per second. How far from the bottom do they pass?
Ans. 96.82ft.
Problem 3., A ball is shot upward with a velocity of 50 feet per second. >
One second later another is shot upward with a velocity of 100 feet per second, v
When and where will they pass and with what relative velocity?
Ans. 1.412 sec. 38.5 ft. 82.2 ft. per sec.
Problem 4. A stone is dropped into a well and 3.2 seconds later the sound
of the splash is heard. What is the depth of the well? (Use 1120 feet per
second as the velocity of sound.) Ans. 151 ft.
Problem 5. A pile-driver hammer weighing 800 Ibs. drops 12 feet upon a
pile. What is the velocity of striking? If the friction of the air and guides is
assumed constant and equal to 50 Ibs., what is the velocity of striking?
Ans. 27.8 ft. per sec. 26.92 ft. per sec.
Problem 6. The hammer of the pile driver in Problem 5 is drawn back up
at a velocity of 4 feet per second. If this velocity is gained in 0.25 of a second
by means of the clutch, what is the tension in the supporting cable?
Ans. 1198 Ibs.
Problem 7. Fig. 221 represents a body weighing 30 pounds which rests
upon a 45° plane, connected by a cord to another weighing 50 pounds which
rests upon a 10° plane. If the coefficient of friction /= 0.25, determine if the
bodies will move. Determine T, FI and F2.
Ans. T = 11.25 Ibs. Ft = 5.3 Ibs. F2 = 12.31 Ibs.
FIG. 222
Problem 8. Two blocks, A weighing 10 Ibs. and B weighing 12 Ibs., Fig.
222, slide down a 30° plane in contact. They start from M with an initial
velocity of 5 feet per second and reach AT, 100 feet from M, with a velocity of
50 feet per second. There is friction under B but none under A. Compute
the coefficient of friction, the pressure between the blocks, the time to move
from M to N and the velocity at the middle.
Ans. / = 0.244. P = 1.15 Ibs. i = 3.635 sec. VM = 35.55 ft. per sec.
Problem 9. A ball is dropped from the ceiling of an elevator 8 feet high.
Find the time to drop to the floor (1) when the elevator is at rest; (2) when it
138
APPLIED MECHANICS
[CHAP, vii
is moving upward with a uniform velocity of 10 feet per second; (3) when it
is being accelerated upward at 10 feet per second per second; (4) when it is
being accelerated downward at 10 feet per second per second; (5) when it is
being accelerated downward at 32.2 feet per second per second.
Ans. (1) 0.704 sec. (3) 0.615 sec. (4) 0.849 sec.
Problem 10. Two blocks are connected by a cord as shown in Fig. 223.
The friction of the cord over the curved surface at A is 10 pounds. The
coefficient of friction under the 100-pound block is 0.1. Compute TI, T2, and
the time required for the block to move 10 feet from rest.
Ans. Ti = 95.6 Ibs. T2 = 105.6 Ibs. *10 = 3.93 sec.
FIG. 223
FIG. 224
Problem 11. In Fig. 224, find the weight W necessary to give the 20-pound
body an acceleration of 12 feet per second per second if the coefficient of
friction under it is 0.4. (Neglect the mass of the pulleys and cord.)
Ans. W = 67 Ibs.
Problem 12. A car starts from rest on a 1 per cent grade and runs down
under the influence of gravity for one mile. From there the track is level.
Train resistance is assumed constant at 12| pounds per ton. Find the velocity
of the car at the lower end of the grade, the time till it comes to rest and the
total distance traveled.
Ans. v = 35.7 ft. per sec. ti = 296 sec. fe = 177.5 sec. s = 8450 ft.
. Problem 13. In the apparatus of Fig. 225 the coefficient of friction under
the blocks is 0.20. Find the acceleration of the blocks, the tension in the cord
and the time to move 20 feet from rest.
Ans. a = 9.22 ft. per sec. per sec. T = 96 Ibs. t = 2.08 sec.
A B
FIG. 225 FIG. 226
Problem 14. The door shown in Fig. 226 weighs 200 pounds and is hung
from a track by means of wheels at A and B. The wheel at A is broken and
slides on the track, / being J. Find the amount of the force P applied as
shown to give the door an acceleration of 8 feet per second per second. Find
the reactions at A and B.
Ans. P = 84.7 Ibs. Ay = 105 Ibs. Ax = 35 Ibs. B = 95 Ibs.
RECTILINEAR MOTION 139
Problem 15. A block 1 foot square and six feet long stands on end on a
carriage with its sides parallel to the direction of motion. The coefficient of
friction/ = £. As the acceleration of the carriage is increased, will the block
tip or slide first, and for what value of the acceleration?
Ans. a = 5.36 ft. per sec. per sec.
Problem 16. A small wooden beam is deflected 1 inch by a weight of
1| pounds. If a weight of 1 pound is placed on the beam and the beam is set
in vibration, find the time of one complete vibration. Ans. T = 0.261 sec.
Problem 17. A weight of 5 pounds is supported by a cantilever beam
spring, which is deflected 0.2 ft. below its neutral position. Find the time of
one vibration. Ans. T = 0.495 sec.
Problem 18. A weight of 10 pounds hung from a spiral spring makes 107
vibrations per minute. What is the scale of the spring?
Ans. 3.25 Ibs. per inch.
Problem 19. What must be the scale of a spring on an 80,000-lb. car so
that if the car strikes a bumping post when moving with a velocity of 3 miles
per hour, the spring will not be compressed more than 2 inches?
Ans. 144,200 Ibs. per inch.
Problem 20. A river is a half mile wide and flows at the rate of 8 miles per
hour. A motor boat which in still water could travel 12 miles per hour is
headed straight across. Where will it strike the opposite bank? What time
will be required? Ans. 1760 ft. below, t = 2 min. 30 sec.
Problem 21. In Problem 20, at what point would the boat strike if headed
30° up-stream? What time would be required?
Ans. 508 ft. below, i = 2 min. 53 sec.
Problem 22. A belt runs crossed between two 18-inch pulleys, 10 feet apart.
Find the relative velocity of the two parts of the belt where they cross when
the rim speed is 10 feet per second.
Ans. 19.77 ft. per sec. in direction of axes of pulleys.
Problem 23. A locomotive drive wheel is 6 feet in diameter and has a
15-inch crank. When the locomotive is running at 60 miles per hour and the
piston is at the forward end of the stroke, what is the absolute velocity of the
crank pin? What is its absolute velocity 90° farther on?
Ans. 95.25 ft. per sec., 22° 40' below hor. 51.35 ft. per sec. hor. forward.
Problem 24. What is the absolute velocity of the top and
bottom points on the rim of the drive wheel of Problem 23?
What is then- velocity relative to the frame of the loco-
motive?
Ans. 176 ft. per sec. 0. 88 ft. per sec. forward. 88 ft. per
sec. backward.
Problem 26. The three bodies, A, B and C, Fig. 227,
weigh 1 pound, 2 pounds and 3 pounds, respectively. If
they are supported in the position shown and then released
simultaneously, what will be the velocity of each body two
seconds later? (Neglect mass of cords and pulleys.)
Ans. Vi = 26.4 ft. per sec. upward. v2 = 19.0 ft. per
Vi = 3.8 ft. per sec. downward.
CHAPTER VIII.
CURVILINEAR MOTION.
91. Velocity in Curvilinear Motion. The velocity of a particle
is its rate of motion, or the time rate of change of its position in
space. If the particle moves along a curved path, its motion is
said to be curvilinear. In this chapter only plane curvilinear
motion will be discussed.
If a particle moves along the curved path ABC, Fig. 228, its
velocity at any point in its path, as at A or B, is in the direction of
the path at that point. The velocity is
therefore changing constantly in direction
and may also be changing in amount. An
approximate value of the velocity of the
particle at any point may be obtained by taking the average velocity
over a small space which includes the point. If the arc AB is the
distance traveled in A£ time, the vector AB is the displacement in
A£ time. Then the ratio -r-r- gives the average velocity of the body
between A and B. If A£ becomes dt, approaching zero as a limit,
point B will approach point A as a limit, so the limiting direction
of the vector AB is the tangent at A, which gives the direction of
the instantaneous velocity at A. As point B approaches point A
the vector AB and the arc AB approach equality. If the length
of the arc AB traversed in dt time is ds, -37 is the magnitude of the
instantaneous velocity, or the speed, at A.
92. Acceleration in Curvilinear Motion. Acceleration is the
time rate of change of velocity. In curvilinear motion the velocity
necessarily changes continually in direction
and may also change in amount. Let the
particle move along the curved path A\A,
Fig. 229 (a), in time A*, and let t;, and v A'
be the instantaneous velocities at Ai and
A, respectively. If from any point 0, Fig.
229(b), vector OBi is laid off equal and parallel to vi, and vector
140
(aj
ART. 93] CURVILINEAR MOTION 141
OB is laid off equal and parallel to v, vector BiB represents the
D r>
total change in velocity, and —rr- gives the average rate of change
of velocity, or acceleration, between ^i and A. The limiting value
of the average acceleration as A£ becomes dt is the instantaneous
acceleration at A.
93. Tangential Acceleration and Normal Acceleration. It is
seen from Art. 92 that in general the acceleration is not in the
direction of the velocity of the particle. Since the direction of the
velocity is usually the direction of reference, the resultant accelera-
tion a is usually determined by means
of its two components, at and an. The
component at is the tangential com-
ponent of the acceleration and is paral-
lel to the direction of the velocity. The
component an is the normal component
of the acceleration and is perpendicular
to the direction of the velocity.
If A\ and A, Fig. 230 (a), are consecu-
tive points in the path of the particle,
the distance A\A being the distance ds which is traversed
in time dt, then BiB, Fig. 230 (b), is the change in velocity
7? 7?
in time dt The average acceleration is -^— and is the instan-
taneous acceleration at A since dt approaches zero as a limit.
7? 7?
The acceleration — j— is resolved into its tangential and normal
DT?
components, -5— = at, parallel to the direction of the velocity at
7? D
A, and — ^— = a», normal to the direction of the velocity at A.
Since A\ and A are consecutive points in the path, DB = v — Vi =
dv, so
dv
It will be seen that this is the rate of change of speed at
point A.
In the limit,
142
APPLIED MECHANICS
[CHAP, vin
Also,
Then
ds
dd = -
an =
, ds
and -7- = v.
at
dt
> _ vdB
~~~dT"
v2
an = —
v ds v2
p dt p
The expressions at = -77 and an — — are important ones and
dt p
should be kept carefully in mind.
From the discussion above and from the principle of effective force,
Art. 85, it follows that if a particle of weight W is moving in a curved
path and at any instant has a tangential acceleration at = -JT and
v2 W
a normal acceleration an = — , the effective force — a must be the
P Q
W dv W v 2
resultant of the two components, Ft = — -77 and Fn =
g dt g p
94. Uniform Motion in a Circle. The discussion in Art. 93
will be understood better by considering its application to several
typical problems. Consider first the case of a block sliding freely
around a smooth groove, the plane of the groove being horizontal.
The top view is shown in Fig. 231 (a). Consider the block as the
free body, as in Fig. 231 (c). The pull of gravity W and the
vertical reaction of the groove R are both normal to the path and
balance each other, since there is no
acceleration in the vertical direction.
Since the block and groove are con-
sidered to be smooth, there is no tan-
gential accelerating force, so the speed
is constant and at = 0. In the direc-
tion of the normal to the path at each
point there is an inward pressure Fn
of the groove upon the block which
turns the block out of the straight path it would otherwise follow.
W W v2
From Art. 93, this normal accelerating force Fn = — an =
y y
Since v and r are constant in amount, the accelerating force Fn is
constant in amount and is always directed toward the center of
the circle.
ART. 94] CURVILINEAR MOTION 143
This normal force which the groove exerts upon the block is
called the constraining or centripetal force and is the resultant of
the external force system acting upon the body. The equal and
opposite inertia force with which the body resists a change of
direction of velocity is called the centrifugal force, since with
respect to the groove the result is the same as if a real force
were acting on the block. This inertia force is the normal
reversed effective force, which, if added to the external system,
will produce static conditions, since in this case there is no tangen-
tial acceleration. In Fig. 231 (c), W and R are equal and balance.
Fn is an unbalanced force and causes acceleration toward the
center. If now, as in Art. 85, the reversed effective force, equal in
W v2
amount to and opposite in direction to the acceleration, be
added to the system acting upon the free body, the body will be
under static conditions and all static equations will apply.
The problem of the conical pendulum illustrates this method
better. In Fig. 232 (a), A is a small weight hung by a cord or light
rod from the point 0 on a vertical axis about
which it rotates in a horizontal plane. If
the speed is constant, the angle 6 is constant
and there is no force in the direction of the
motion. In Fig. 232(b), the weight is shown
as a free body with all the external forces
acting upon it, and in addition the reversed
Wv2
effective force , acting opposite to the ac-
celeration. The problem is now one in statics, and the equations
2>FX = 0 and 2FV = 0 determine the unknown quantities. The
moment equation, £ M0 = 0, gives 6 and h in terms of the velocity.
g r
~ h~ gr ~ v2'
EXAMPLE 1.
In Fig. 232, let I = 2 feet and W = 5 pounds. What is the speed of the
weight if 6 = 30°? What is the tension in OA?
Solution: — If 0 = 30° and I = 2 ft., r = 1 ft.
144
APPLIED MECHANICS
[CHAP, vin
Equation 2Fy = 0 gives
T X 0.866 = 5.
T = 5.771bs.
Equation *LFX = 0 gives
v = 4.31 ft. per sec.
Or, by using equation SM0 = 0,
K 7,2
* 1.732 -5 XI.
v = 4.31 ft. per sec. as before.
EXAMPLE 2.
Find the superelevation e of the outer rail of a railroad track of gage G on
a curve of radius r so that there is no resultant flange pressure when the car
has a speed of v feet per second.
Solution: — In Fig. 233 (a), Ri and R2 are the pressures of the rails on the.
wheels; R is their resultant; W is the weight of
the car and r is the radius of curvature of the track.
Since r is horizontal and the car is accelerated toward
the center, the reversed effective force must
9 r
act horizontally through the center of gravity. The
conditions are now static conditions, so the three
forces must be in equilibrium and can be made to
form a closed triangle of forces, Fig. 233 (b) . Then
^L vl + w = tanfl, or tan0 = — •
9 r 9r
For small angles the sine and the tangent are
approximately equal, so from Fig. 233 (c),
gr
If 9 becomes large, so that the tangent and the sine cannot be considered
equal, the superelevation e is determined from e = G sine.
Problem 1. With dimensions and weight the same as in Example 1 above,
find the speed v necessary to keep the cord at an angle of 45°. What is the
corresponding tension T? Solve also for 60°.
Ans. v = 6.75 ft. per sec. T = 7.07 Ibs. v = 9. 82 ft. per sec. T = 10 Ibs. ,
Problem 2. Determine the superelevation of the outer rail of a track of
gage G = 4.9 feet (center to center of rail) on a curve of radius r = 2865 feet
to give zero resultant flange pressure at a speed of 30 miles per hour.
Ans. e = 0.103 ft.
Problem 3. If a 100,000-lb. car has a speed of 60 miles per hour on the
curve with superelevation as determined in Problem 2, what is the resultant
flange pressure parallel to the ties? If the car is at rest on the curve, what is
the resultant flange pressure parallel to the ties? Ans. 6290 Ibs. 2100 Ibs.
ART. 95]
CURVILINEAR MOTION
145
95. Simple Circular Pendulum. A simple circular pendulum
consists of a particle vibrating in the arc of a vertical circle under
the influence of gravity and some constraining radial force. The
ideal simple circular pendulum may be closely approximated by
means of a small heavy sphere at the end of a light cord.
Let A, Fig. 234, be such a body suspended by cord OA, of length
/, and let distance along the arc be measured
from C, positive to the right. The only
force in the direction of motion is — W sin 6.
From F = — a,
at = —gsm6.
vdv = ads = —g sin 0 ds.
The expression for t in terms of the integral
of sin Ods is a complicated elliptic form, but
an approximate solution is comparatively
simple and for vibrations of small amplitude is very slightly in
error. For small values of 6, sin 0 = 6, approximately, so the
equations above become
FIG. 234
«- ~dO=-j
since 0 =
and
v dv = — j-sds
Let v0 be the velocity at C and v the velocity at A.
Then fvdv = -f / *sds.
*J va I *JQ
I
Since v = 0 when s = SB,
The insertion of this value of VQ* in the equation above gives
V =
146 APPLIED MECHANICS [CHAP, vni
Since
g
If time is measured from the instant the pendulum is at C, moving
to the right, this becomes
ds
- o2
9 SB
To get the time required for the pendulum to move from C to B,
let s = SB- Then
f™
To get the time required for the pendulum to move from C to B
and back to C, let s = 0. Then
tc =
The time required for the pendulum to move from C to B is there-
fore the same as that to move from B to C.
Motion to the left of C exactly corresponds to motion to the
right of C, so the time of one complete period of vibration,
g
This equation for T is independent of 6, so the time of vibration is
independent of the amplitude for small values of 0.
It will be seen from the equation at — — ? s, that for vibrations
l
of small amplitude the acceleration is proportional to the displace-
ment and so the motion is practically simple harmonic motion.
Since the ball of the pendulum is accelerated toward the center
v2
with an acceleration -=• , the summation of forces normal to the path
I
gives
W v2
— j,
ART. 96] CURVILINEAR MOTION 147
P being the tension in the cord.
P = Wcose + — T-
9 I
Problem 1. What is the length I of a simple pendulum which will vibrate
from one side to the other (B' to B, Fig. 234) in 1 second? Ans. 2.26 ft.
Problem 2. A mine cage is suspended from a cable 500 feet long. What
is the time of one complete oscillation? Ans. 24.75 sec.
Problem 3. A girder weighing 800 pounds is suspended from a cable 40
feet long. What horizontal force is necessary to pull it 6 feet out of its vertical
position? What is the tension in the cable as the girder is allowed to swing
back through its vertical position? Ans. 121 Ibs. 818 Ibs.
96. Velocity of a Body in a Vertical Curve. In Fig. 235, let
A be a body of weight W moving along a
smooth vertical curve under the influence
of gravity, and let 6 be the angle between
the horizontal and the tangent to the curve
at any point. The reaction N of the
smooth surface is normal to the path, so if JPIG 235
the resistance of the air is neglected the
only force in the direction of the motion is — W sin 6, as in Art.
F
95, and a = -^ = —g sin 6.
vdv = ads = —g sin d ds.
sin 0 ds = dy,
so vdv = —gdy.
Let VQ be the velocity at point B. Then
Xv fy
vdv = —g I dy.
j *J h
v2- v02= 2g(h-y).
It will be seen from this equation that if a body moves along a
frictionless path under the influence of gravity, the change in speed
is the same as if it fell freely through the vertical distance between
the two points. (See Equation 3, Art. 83.)
As in Art. 95, the normal constraining force P is given by
W i?
P = ±W cos8+ - ->
Q r
r being the radius of curvature of the path. The negative sign
148
APPLIED MECHANICS
[CHAP, vin
is used if the body is above a horizontal line through the center
of curvature.
EXAMPLE.
A body B weighing 2 pounds is rotating in a vertical circle at the end of a
cord 1 foot long, as shown in Fig. 236. If its
velocity at the bottom point is 15 feet per second,
what is its velocity at the top point Al What
is the tension T in the cord at that point? What
is the least velocity it can have at point A which
will keep it in its circular path?
Solution: — If the X axis is taken through B,
the value of h in the equation above becomes zero.
Then
—, VAZ = v0z — 2 gy = 225 - 128.8.
Vo= /O **
FIG. 236 y/=96.2
VA = 9.81 ft. per sec.
When the body is at point A, the only forces acting upon it are its weight
and the tension in the cord, both downward. These produce the normal
acceleration an = —
r
96.2 ft. per sec. per sec.
W
From the equation F = — a,
2 =
X 96.2.
32.2
T = 3.98 Ibs.
Since the cord cannot have a compressive stress, the minimum value T can
W v2
have is zero. When T is zero, equation F = -- gives
v - 5.67 ft. per sec.
This is the least velocity which will keep the body in its circular path.
Problem 1. If the body in Fig. 236 starts from rest at point C, what is its
velocity as it passes the 45° point? What is its velocity at the bottom point?
Am. 6.75 ft. per sec. 8.03 ft. per sec.
Problem 2. If a body weighing 10 pounds is rotating in a vertical circle
at the end of a cord 10 feet long, what is the least velocity at the bottom which
will keep it in the circular path at the top? What is the tension in the cord
when the body is at the bottom? When it is 60° from the top? When it is
30° from the bottom? Ans. 40.1 ft. per sec. 60 Ibs. 15 Ibs. 56 Ibs.
Problem 3. A body is thrown from a cliff 2000 feet high with a velocity of
200 feet per second. What will be the amount of its velocity as it strikes the
surface of the water below, if the air resistance is neglected?
Ans. 411 ft. per sec.
ART. 97]
CURVILINEAR MOTION
149
Problem 4. If in Fig. 237 the radius of the circle on the " loop the loop " is
6 feet, what must be the height h in order that the car will just pass the point
A without leaving the track? Neglect frictional resistance.
Ans. h = 15 ft.
FIG. 237
97. Motion of Projectile, Air Resistance Neglected. If a body
is impressed with an initial velocity and then moves freely through
the air under the influence of gravity, it is called a projectile. In
this discussion the resistance of the air will be neglected. For
projectiles with high velocities the error is considerable.
Let the projectile be discharged with an initial velocity of VQ at
an angle a with the horizontal, as shown in Fig. 238. The initial
velocity VQ may be resolved into its horizontal and vertical com-
ponents, ^o cos a and VQ sin a, respectively. Since there is no
horizontal force acting upon the body, the horizontal velocity
remains unchanged throughout its motion.
a* = 0, vx = v0 cos a,
x = v0t cos a.
In the vertical direction the force of gravity is acting continuously,
so the vertical component of the velocity is the same as that of a
body projected upward with an initial velocity of VQ sin a. See
Art. 83.
av = — g, Vj, = v0 sin a - gt,
y = Votsina — \ gt2.
150 APPLIED MECHANICS [CHAP, vm
To determine the time till the projectile reaches the highest
point in its path, let vy = 0; then t = - — . The greatest
J/
height h will be found by putting this value of t in the expression
for y. This gives
. v02 sin2 a
Tf
For varying values of a, the maximum value of h is evidently
given when sin2 a = 1, so a for maximum h = 90°.
The time the projectile is in the air until it reaches the same
level again is obtained by letting y = 0. This gives
_ 2 VQ sin a
t — ~ >
which is twice the time required for it to reach the top point in its
path.
The range d, the horizontal distance through which it travels
until it reaches the same level again, is given by the product of
the time t just obtained and the constant horizontal velocity^
, 2 v02 cos a sin a v02 . 0
d = = — sin 2 a.
g g
For varying values of a it is evident that d will be a maximum
when sin 2 a = 1. Then 2 a = 90° and a = 45°.
The amount of the velocity at any point in the path is given by
v = Vvx2 + vy2 = Vv02 - 2 gy.
The direction of the velocity at any point in the path is the same
as the tangent to the path, and is given by
6 being the angle the tangent to the curve makes with the hori-
zontal.
The equation of the path in terms of x and y is obtained by
eliminating t between the equations for x and y.
gx2
This is the equation of a parabola with its axis vertical.
ART. 97] CURVILINEAR MOTION 151
EXAMPLE 1.
A projectile is discharged with a velocity of 200 feet per second at an angle
of 30° with the horizontal. Determine the maximum height, the time until
it returns to the same level, and the range. Determine also the velocity at
the end of 2 seconds.
Solution: — The vertical component of the velocity is VQ sin a =• 100 ft.
per sec. The horizontal component of the velocity is VQ cos a = 173.2 ft. per
sec.
, Vo2sin2a 100X100
The time until it reaches the same level again is 2 V° Sm a = ^ = 6.21 sec.
g 6Z.Z
The range d is given by the product of the horizontal velocity and the time.
d = 173.2 X 6.21 = 1075ft.
The vertical component of the velocity at the end of 2 seconds is given by
vy = VQ sin a — gt = 100 — 64.4 = 35.6 ft. per sec.
The horizontal velocity is constant, 173.2 ft. per sec. The resultant velocity
v = Vvx2 + vv2 = 176.8 ft. per sec. The angle with the horizontal is given
by its tangent
EXAMPLE 2.
From a tower 200 feet high a stone is thrown downward at an angle of 45°
with the horizontal, with a velocity of 80 feet per second. How far away from
the tower does it strike the ground and what is the time required?
Solution: — The horizontal component of the velocity is VQ cos a = 56.56
ft. per sec. The vertical component of the velocity is v0 sin a = 56.56 ft. per
sec. downward. The vertical motion is the same as that of a body projected
vertically downward with an initial velocity of 56.56 ft. per sec. From equa-
tion (2), Art. 83,
200 = 56.56 1 + 16.1 1\
t = 2. 18 sec.
Since the horizontal velocity is constant the horizontal distance traveled in
2.18 seconds is
xi =56.56 X2.18 = 123.4ft.
Problem 1. A shot is fired at an angle of 60° with the horizontal, with a
velocity of 500 feet per second. Find the height and the range.
Ana. h =2911 ft. d = 6725 ft.
Problem 2. What is the maximum height and the maximum range theo-
retically possible for a projectile with a muzzle velocity of 2000 feet per second?
Ans. h = 11.78 miles, d = 23.56 miles.
Problem 3. A gun with a barrel 3 feet long shoots a bullet with a muzzle
velocity of 600 feet per second. When shooting at a target 1000 feet distant,
to what height should the rear sight be raised? Ans. 1.615 inches.
152 APPLIED MECHANICS [CHAP, vm
GENERAL PROBLEMS.
Problem 1. A 10-pound weight at the end of a 3-foot cord revolves about
a vertical axis as a conical pendulum. What is the tension in the cord and the
speed of rotation if 6 = 30°? Ans. T = 11.55 Ibs. v = 5.28 ft. per sec.
Problem 2. A cast iron governor ball 4 inches in diameter has an arm
15 inches long. Neglecting the weight of the arm, find the tension and the
speed if 0 = 75°. Ans. T = 33.76 Ibs. v = 12.04 ft. per sec.
Problem 3. A car is moving on a horizontal track around a curve of 600
feet radius with a speed of 30 miles per hour. A weight of 40 pounds is sus-
pended from the ceiling by a cord 3 feet long. What is the tension in the cord,
the angle with the vertical and the horizontal displacement?
Ans. 40.2 Ibs. 5° 42'. 3.6 inches.
Problem 4. What superelevation of the outer rail would be necessary in
Problem 3 in order to make the weight hang parallel to the sides of the car?
Ans. 5.88 in.
Problem 5. A standard gage interurban railway track has a curve of 100
feet radius. What should be the superelevation of the outer rail so that a car
moving with a speed of 10 miles per hour will have zero resultant flange
pressure? Ans. 3.92 in.
Problem 6. In a fog, a car whose center of gravity was 6 feet above the
track struck the curve of the preceding problem while moving with a speed of
30 miles per hour. What happened? (Solve graphically.)
Problem 7. At what angle must an automobile speedway be banked on a
curve of 250 feet radius for a speed of 60 miles per hour so that there is no side
thrust on the wheels? Ans. 43° 50'.
Problem 8. If the coefficient of friction between the wheels and the track
in Problem 7 is ^, to what angle can the banking be reduced before the wheels
skid on the curve? Ans. 25° 30'.
Problem 9. A common swing 20 feet high is designed for a static load of
400 pounds with a factor of safety of 4. If two boys, each weighing 100 pounds,
swing up to the horizontal on each side, what is the factor of safety?
Ans. 2f.
Problem 10. If the swing of Problem 9 is vibrating 60° on each side of the
vertical, what will be the total tension in the supporting ropes?
Ans. 400 Ibs.
Problem 11. If the swing of Problem 9 vibrates through an angle of 10°.
how many complete vibrations are made per minute? Ans, 12.12.
Problem 12. A ball at the end of a cord 2 feet long is swinging in a complete
vertical circle with just enough velocity to keep it in the circle at the top. If
it is released from the cord when it is at the top point, where will it strike the
ground 4 feet below the center of the circle? Ans. 4.89 feet away.
Problem 13. If the ball of Problem 12 is released 45° later, where will it
strike the ground? Ans. 4.27 feet from vertical through center.
Problem 14. The muzzle velocity of a projectile is 1200 feet per second
and the distance of the target is 3 miles. What must be the angle of elevation
of the gun? Ans. 10° 22'.
Problem 15. From a car moving with a speed of 60 miles per hour a stone
is thrown horizontally at right angles to the direction of the car with a velocity
CURVILINEAR MOTION
153
of 100 feet per second. Where will it strike the ground and with what velocity
if the car is 10 feet above the level of the ground?
Ans. 78.8 ft. from track, 69.4 ft. forward, v = 135.6 ft. per sec.
Problem 16. A small block slides freely down the quadrant shown in
Fig. 239. Determine the distance x\, the equation of the path BC and the
velocity at C.
Ans. xi = 10.95 ft. y = -0.0834 x2. vc = 28.9 ft. per sec.
Problem 17. While blasting out a concrete foundation the most distant
pieces fell 800 feet away. What was their initial velocity?
Ans. 160 ft. per sec.
FIG. 239
FIG. 240
Problem 18. If a target is 1000 feet distant horizontally and is 200 feet
higher than the gun, what angle of elevation is necessary if the muzzle velocity
is 800 feet per second? Ans. a = 12° 50'.
Problem 19. A block starts from rest at the top of a sphere 4 feet in diam-
eter, Fig. 240, and slides without friction to point 0' where it leaves the surface
of the sphere. Locate this point. Find also the equation of the path of the
block after leaving the surface, and the distance from the sphere to the point
where the block strikes the plane upon which the sphere is resting.
Ans. 6 = 48° 10'. y = 1.118 x - 0.843 x2. a + Xi = 2.92 ft.
k- 10' - '
FIG. 241
Problem 20. The height of the starting point for the car on the " loop the
loop," Fig. 241, is 30 feet above the bottom of the loop. What is the velocity
of the car at the top of the loop? What is the pressure against the top if the
man and car weigh 300 pounds? Ans. 35.85 ft. per sec. 2100 Ibs.
Problem 21. In Problem 20, if the track from the bottom of the loop to the
right rises 2 feet in a distance of 10 feet, how wide a gap, mn, can be leaped?
Ans. 21.4 ft.
CHAPTER IX.
ROTATION.
98. Angular Displacement. If a particle describes a curvilin-
ear motion with a constant radius r, the motion is called rotation
and the angle described by the radius is called angular displace-
ment. The unit of angular displacement is the radian. The
radian is the angle at the center subtended by an arc equal in
length to the radius. In Fig. 242, the length of the arc AB is equal
to the radius r, so the angle AOB is one radian.
Let the length of the arc ABC be s. Since any
angle is proportional to its subtending arc,
FIG. 242 s = r-
There are 2?r radians subtended by a complete circumference.
Hence 2 TT radians = 360° and 1 radian = - = 57°.3. (More
ZTT
accurately 57°.29578.)
Angular displacement in the counterclockwise direction is
considered to be positive, that in the clockwise direction
negative.
Problem 1. Reduce to radians: 45°; 100°; 900°; 4 revolutions.
Ans. 0.7854. 1.746. 15.71 25.13.
Problem 2. Reduce to degrees: 2 radians; ir radians.
99. Angular Velocity. Angular velocity is the time rate of
angular displacement. If equal angular displacements occur in
equal intervals of time, the angular velocity is constant. Let o>
represent the angular velocity in radians per second. Then if 6
represents the angular displacement in time t, the rate of angular
displacement, or angular velocity, is given by
e
154
ART. 100] ROTATION 155
If the angular velocity varies, the average angular velocity for
any small interval of time A£ is given by
»M.
The instantaneous angular velocity is given by
de
w = 5'
Angular velocity is commonly given in revolutions per minute
(r.p.m.), which must usually be reduced to radians per second for
the solution of problems.
1 27T
1 r.p.m. = 77~ rev. per sec. = -^ rad. per sec.
Since angular velocity involves only magnitude and direction, it
is a vector quantity and may be represented graphically by a
vector. In the same way as for vectors of couples, the vector is
drawn parallel to the axis of rotation with the arrow pointing in the
direction from which the rotation appears counterclockwise or
positive. Vectors representing angular velocity may be combined
graphically into their resultant vector which represents the result-
ant velocity. Conversely, the vector representing an angular
velocity may be resolved into component vectors.
The relation between the linear and the angular velocities of a
point moving in a circular path is obtained as follows.
ds ~. rdd •„ , dO
By definition, v = -r . Since s = r6, v = -5- . But -r = co, so
v = rco.
Problem 1. A pulley 40 inches in diameter rotates at 120 r.p.m. What is
its angular velocity in radians per second and what is the speed of the belt in
feet per second? Ans. 12.57 rad. per sec. 20.94 ft. per sec.
Problem 2. The smaller of two friction wheels is 3 inches in diameter and
the larger is 24 inches in diameter. If the smaller wheel rotates at 30 r.p.m.,
what is the angular velocity of the larger wheel? What is the speed of the rim?
Ans. co = 0.3927 rad. per sec. v = 0.3927 ft. per sec.
100. Angular Acceleration. Angular acceleration is the time
rate of change of angular velocity. If the angular velocity is con-
stant, the angular acceleration is zero. If the angular velocity
changes by equal amounts in equal time intervals, the angular
acceleration is constant. If the angular velocity changes by
156 APPLIED MECHANICS [CHAP, ix
unequal amounts in equal time intervals, the angular acceleration
is variable.
If the angular acceleration is constant, its value may be ob-
tained by dividing the total change in angular velocity by the
time t in which the change was made. If a. represents the angular
acceleration and co the change in angular velocity,
CO
a"F
If the angular acceleration is variable, its instantaneous value at
any point is given by
_dco _dM
~ dt~ dP
By eliminating dt from the two equations, co = "^and a = -^->
there is obtained the equation
co dco a d6.
The unit of angular acceleration is the radian per second per
second. Like angular velocity, angular acceleration is a vector
quantity and may be represented graphically by a vector. Angular
acceleration has sign, counterclockwise acceleration being usually
taken as positive and clockwise acceleration negative.
The relation between tangential and angular acceleration is
obtained as follows. By Art. 93, at = -r.' Since v = rco, dv = r dco
r da) ~. da)
and at = —rr- Since -77 = a,
a.t= ra.
v2
By Art. 93, an = — • Since v = rco,
a« = rco2.
It is seen that the tangential acceleration varies directly with the
radius and with the angular acceleration. The normal accelera-
tion varies with the radius and with the square of the angular
velocity. It is independent of the angular acceleration.
101. Simple Harmonic Motion: Auxiliary Circle Method.
The equations of simple harmonic motion may be derived very
easily by means of the Auxiliary Circle. If a point describes a
circular motion with uniform speed, the motion of its projection
upon a diameter is simple harmonic motion, as will be shown later.
ART. 101] ROTATION 157
Let P, Fig. 243, be the point, moving counterclockwise at 01
radians per second with an angular displacement of 6 from A. If
time is measured from the instant when
P is at point A, 6 = ut. Let the tan-
gential velocity of the point P be t'0.
Let P' be the projection of P upon the
horizontal diameter. The velocity of
P' at 0 is evidently the same as that
of P at A, since both are moving hori-
zontally. At any point in its path,
the velocity v of P' is equal to the
horizontal component of the velocity
of P in its corresponding position, or
V = V0 COS (0t.
Since the point P is moving around the circle with uniform speed,
its only acceleration is toward the center and is an = ro>2, repre-
sented by the vector PC. The acceleration a of the point P' is
along its path and is equal to the horizontal component PE of the
acceleration of point P. This horizontal component is equal to
an sin 6 = rco2 X - = co2s. The direction is negative, so
a = -co2s.
Since co2 is a constant, a is proportional to the displacement s, hence
the motion of P' is simple harmonic motion.
By squaring the first equation above,
V2 = V<? COS2 wt.
s2
Since cos2 otf = 1 — sin2 cot = 1 -- -^
- Sl\
V = \/V02 -
From Fig. 243,
s = r sin cot.
Solving for t,
1 . . s
t = - sin"1 - •
<o r
In these equations, t is the time from position 0. From the last
equation the time from 0 to B is evidently
1 TT
t = - sin-1! = - —
0) <60>
158
APPLIED MECHANICS
[CHAP, ix
If K is substituted for o>2, the formulas of this article become
identical with those of Art. 87, and as was there shown, the period
is four times the time from 0 to B.
CO
If the crank pin of a reciprocating engine rotated with uniform
speed, the piston would have simple harmonic motion if the con-
necting rod were of infinite length, or if the slotted slider shown in
Fig. 244 were used.
\
FIG. 244
EXAMPLE.
In the mechanism shown in Fig. 244, let r = 1 foot and let the crank pin
be rotating at 120 r.p.m. Determine co, T, v0 and the maximum acceleration
of the piston. If the weight of the piston and slider is 200 pounds and the
steam pressure F is zero, what is the pressure FI of the crank pin on the slider
when the piston is at its end piston?
Solution: — 120 r.p.m. = 2 rev. per sec. = 4ir rad. per sec.
w = 4?r = 12.57 rad. per sec.
.
2
v0 =
= 12.57 ft. per sec.
Maximum a = —u?r = —158ft. per sec. per sec.
At the end position the only force acting upon the piston and slider in the
direction of its motion is the pressure of the crank pin FI. The acceleration is
toward the left, so the force FI must act toward the left. From F = Ma,
nf\f\
! 158 = 981 Ibs.
Problem 1. In Fig. 244 let the angle POA of the crank with the vertical
radius OA be 30° and let all other data be the same as in the Example above.
Determine the velocity and the acceleration of the piston. Determine the time
since it was in its middle position. If the steam pressure F = 500 pounds,
what is the crank pin pressure Fit
Ans. v = 10.88 ft. per sec. a = — 79 ft. per sec. per sec. t = ^ sec.
Ft = 991 Ibs.
ART. 102] ROTATION 159
Problem 2. With the same data as in the Example above, determine the
velocity and acceleration of the piston and its distance from the right end of
the cylinder 0.15 second after passing the middle position moving to the right.
If the steam pressure is 1500 pounds (to the left), what is the crank pin pressure?
Ans. v = 3.88 ft. per sec. a = 150.3 ft. per sec. per sec. Fi = 566 Ibs.
to right.
102. Constant Angular Acceleration. In Art. 82 the following
expressions were derived for the linear motion of a particle with
constant acceleration.
v — v0 -\- at. (1)
s = i (v + VQ) t = vQt + | at2. (2)
t? = t,02 + 2 as. (3)
At any instant these equations apply also to the tangential motion
of a particle moving in a circle with constant tangential accelera-
tion. In the formulas above, a becomes at. From Arts. 98, 99
and 100,
s v at
u = - , w = - and a = —
r' r r
If equations (1) and (2) are divided by r and simplified, they be-
come
a) = <oo + at. (4)
6 = J (<o + coo) t = coot + J at2. (5)
Similarly, if equation (3) is divided by r2, it becomes
(o2 = coo2 + 2 0,9. (6)
Problem 1. A flywheel is brought from rest up to a speed of 60 r.p.m. in
\ minute. What is the average angular acceleration a, and the number of
revolutions required? What is the velocity at the end of 15 seconds?
Ans. a = 0.2094 rad. per sec. per sec. 15 rev. o>i5 = 3.14 rad. per sec. - ^-
Problem 2. If the flywheel of Problem 1 is 12 feet in diameter, what is the
tangential velocity and acceleration of a point on the rim? What is the nor-
mal acceleration at the instant full speed is attained?
Ans. v = 37.7 ft. per sec. at = 1.256 ft. per sec. per sec. an = 237 ft. per
sec. per sec.
Problem 3. A pulley which is rotating at 120 r.p.m. comes to rest under the
action of friction in 3 minutes. What is the angular acceleration and the total
number of revolutions made?
Ans. a = 0.0698 rad. per sec. per sec. 180 rev.
Problem 4. The rim of a 33-inch wheel on a brake shoe testing machine
has a speed of 60 miles per hour when the brake is dropped. It comes to rest
when the rim has traveled a tangential distance of 440 feet. What is the
angular acceleration and the number of revolutions?
Ans. a = 6.4 rad. per sec. per sec. 50.9 rev.
160 APPLIED MECHANICS [CHAP, ix
103. Variable Angular Acceleration. If in a circular motion
the angular acceleration a is variable, its law of variation must be
known in order to obtain the equations of motion. In certain
physical problems the acceleration is zero in the mid-position and
increases directly with the angular displacement 6 to either side
of this position and is oppositely directed. Then a = —KB, K
being a constant. Let the angular velocity at the mid-position be
COQ. Since co du = a dd,
£re
wdco = -K I
Jo
<o = V(o02 - K6
Since o> = -7:, this equation becomes
j/)
dt =
If time is measured from the instant the body is in the mid-
position moving positively, the limits of t are 0 and t, and of 6 are
0 and 6.
de
'o
i ,*VK
t = -7= sin-1
vK <«>o
By transposing,
sin V#Z.
From the above equations it will be seen that the motion is periodic.
(See Equation (2), Art. 87.) These equations apply to the motion
of a torsion balance and approximately to that of a simple pendu-
lum with vibrations of small amplitude.
Problem 1. The balance wheel of a watch is \ inch in diameter and oscil-
lates 45° to either side of its mid-position. The time of one complete oscilla-
tion is \ second. What is its greatest angular acceleration, its greatest angular
velocity, its greatest tangential acceleration and its greatest normal accelera-
tion?
Ans. a = 124 rad. per sec. per sec. coo = 9-87 rad. per sec. at — 2.58 ft.
per sec. per sec. an = 2.03 ft. per sec. per sec.
104. Effective Forces on a Rotating Body. Let Fig. 245
represent any rotating body and P any particle of mass dM, at a
ART. 105] ROTATION 161
radial distance p from the axis of" rotation, 0. Then if the body
has an angular velocity co and an angular acceleration a, the tan-
gential and normal components of the accel-
eration a are at = pa and an = po>2 respectively
for each dM . The effective force for each
dM is given by its two components, dMpa
tangential to its path and d/kfpco2 normal to
its path. F10- 24s
It will be seen that for equal dM's, these tangential and normal
effective forces vary with the radius p, both in amount and
direction. The tangential effective forces vary directly with p
and are always at right angles to it in the direction of the angular
acceleration. The normal effective forces vary directly with p
and are always directed along the radius toward the axis.
105. Moment of Tangential Effective Forces. In Fig. 246 let
0 be the axis of rotation, C the center of gravity, F the resultant
of all the external forces except the reac-
tion at the axis 0, and d the distance from
the axis of rotation to the line of action of
force F. P is any particle of mass dM,
whose tangential effective for"ce is dMpa
FIG. 246 as shown. (Since rotation alone is being
considered, forces parallel to the axis of rotation are neglected.)
By the principle of Art. 85 the impressed forces would be in
equilibrium with the effective forces reversed. Since neither the
normal effective forces (not shown) nor the reaction at 0 (not
shown) have any moment about the axis 0, the moment of the tan-
gential effective forces must be equal to the moment of the im-
pressed forces, or
Fd = CdMfa.
Since at any instant all particles of the body have the same value
of a
Fd = aCdMp*.
The value of / dMp2 if integrated between the proper limits is /,
the moment of inertia of the body with respect to the axis of
rotation. Then
Fd = la.
162 APPLIED MECHANICS [CHAP, ix
The analogy of this equation to the equation F = Ma was dis-
cussed in the footnote to Art. 66.
It is convenient sometimes to consider the moment la as equiv-
alent to the moment of a single force Mra (the total mass X tan-
gential acceleration of mass center). Its moment arm is given by
la = MWa _ fc*
Mra Mra r
Then a force Mra acting tangentially through point Q, Fig. 246,
k2
distant — from the axis has a moment about the axis equivalent
to that of the actual tangential system of effective forces. It is
evident from the discussion above that if the effective tangential
force Mr a were reversed in direction and made to act tangentially
through point Q, its moment would balance the accelerating mo-
ment of the external system of forces and would produce a static
condition of rotation.
If a body rotates about its center of gravity so that points 0
K2
and C coincide, r = 0, so — = infinity. This shows that in this
case the equivalent moment can be given only by a couple of
moment la, and not by a single force.
EXAMPLE 1.
A slender rod of length Z is released from a horizontal position and allowed
to rotate under the influence of gravity alone about a horizontal axis through
one end, perpendicular to the axis of the rod. Determine the amount and
position of the tangential effective force Mra at the instant of starting.
Solution: — Let W be the weight of the rod. From the equation
Fd = 7«,
Wl WP
*L = l- -^ L - ?.
r ~'3 : 2~3
EXAMPLE 2.
Fig. 247 represents a cast iron cylinder 3 feet in diameter and 6 inches
thick, free to rotate about its geometric axis 0. If a force of 10 pounds is
ART. 105] ROTATION 163
applied to a cord wrapped around the cylinder, what is
the angular acceleration, the tangential acceleration, the
angular velocity after 5 seconds and the number of
revolutions it has turned? Neglect the axle friction.
Solution: — The forces acting upon the cylinder are
its weight, the reactions of the supports at 0, and the
force F. The first two forces have no moment about the
geometric axis, so the moment FIG. 247
Fd = 10 X 1.5 = 15ft.lbs.
/o of the cylinder = \ Mr2 = 55.6.
Fd = la.
15 = 55.6 a.
a = 0.27 rad. per sec. per sec.
at = ra. = 0.405 ft. per sec. per sec.
co = at, so cos = 1.35 rad. per sec.
0 = i op = 3.375 rad. in 5 sec.
3.375 -^ 2 TT = 0.537 rev. in 5 sec.
EXAMPLE 3.
Instead of a force F = 10 Ibs., let a weight of 10 Ibs. be hung from the cord
in Example 2. Determine the angular acceleration and the tension T in the
cord.
Solution: — First, let the cylinder be considered as the free body.
The equation Fd = la gives T X 1.5 = 55.6 a.
This equation contains two unknown quantities, so the equation of motion of
the suspended weight must be written.
"-*-««
Since a of the weight = at of the rim of the cylinder, and at = ra,
a = 0.266 rad. per sec. per sec.
T = 9.86 Ibs.
Problem 1. A steel cylinder 2 feet in diameter and 2 inches thick is fastened
firmly to an axle 2 inches in diameter. A weight of 50 pounds is hung from a
cord wrapped around the axle. Find a and T, neglecting the mass of the pro-
truding parts of the axle and also the axle friction.
Ans. a = 1.045 rad. per sec. per sec. T = 49.85 Ibs.
Problem 2. A cast iron flywheel 4 feet in diameter has a rim 2 inches thick
and 12 inches wide. To its axle is fastened a 20-inch pulley around which a
cord passes. If a 100-pound weight is hanging from the end of the cord, what
is the angular acceleration of the flywheel and what is the tension in the cord?
Neglect the mass of the axle, hub, spokes and small pulley, also axle friction.
Ans. a = 0.785 rad. per sec. per sec. T = 97.9 Ibs.
164
APPLIED MECHANICS
[CHAP, ix
Problem 3. A cast iron cylinder 3 feet in diameter and 2 feet long is
rotating at 360 r.p.m. A brake which rubs against the curved surface of the
cylinder has a normal pressure on it of 500 pounds. The coefficient of friction
between the brake and the cylinder is / = 0.2. If the friction on the shaft
is neglected, what is the time required for the brake to bring the cylinder to
rest? Through how many revolutions will it turn?
Ans. 56 sec. 168 rev.
106. Resultant of Normal Effective Forces. As stated in
Art. 104 the normal effective forces for the particles of a rotating
body at any instant are directly proportional to their radii and act
toward the axis of rotation. (See Fig. 245.) In general, the re-
sultant of these normal forces for the whole body is a force and a
couple, as was shown in Art. 39. The solution of this problem in
the general case is involved and difficult, for usually the resultant
force does not act through the center of gravity, and the resultant
couple is hard to obtain since it involves the product of inertia of
the body. Fortunately, nearly all engineering problems in rota-
tion come under a few special cases in which the value of the
couple is zero or is easily obtained and the resultant force is
easily located. Three of these cases will be discussed.
Case 1. // a body has a plane of symmetry and rotates about any
axis normal to this plane, the resultant normal effective force acts
radially through the center of gravity and is equal to Mrco2.
N
Let Fig. 248 (a) represent a body which has a plane of symmetry
QMNP. Let the axis OZ normal to the plane of symmetry be the
axis of rotation, and let the X axis be taken through the center of
gravity C. Let AB be any elementary prism of mass dM, par-
allel to the axis OZ. Since each part of the prism has the same
normal acceleration an = pco2, the resultant of the normal effective
forces for the prism AB is dMpu? acting in the plane of symmetry
toward the axis OZ. Fig. 248 (b) shows the section cut by the
plane of symmetry. The X component of the force dMpw2 is
ART. 106]
ROTATION
165
dMpa>2cosO = dMpw2- = dMu2x, and for the entire body 2FX =
to* / dMx = Mxu2 = Mrw2. The Y component of the force dMpw2
is dMpu2 sin B = dMpu2 - = dMu>2y, and for the entire body, 2FV =
all
co2 / dMy = Myu2 = 0, since y = 0. Hence the resultant of
the normal effective forces is Mrco2 acting in the plane of sym-
metry parallel to the X axis. Since the normal effective forces all
pass through the axis OZ they have no moment about OZ', hence
their resultant can have no moment about OZ and must therefore
lie in the axis OX through the center of gravity.
If the axis OZ passes through the center of gravity, ? = 0, so
the resultant normal effective force Mrco2 = 0.
Case 2. // a body has a line of symmetry and rotates about an
axis parallel to this line, the resultant normal effective force acts
through the center of gravity and is equal to Mru2.
In Fig. 249, let AB be a line of symmetry of the body shown and
let the body be rotating about axis OZ parallel to AB. Consider
the plate EF of thickness dz and mass dM,
whose plane is normal to the axis OZ. By
Case 1, the resultant normal effective force
on plate EF is dMra2, acting through the
center of gravity of the plate toward the axis
OZ. On each similar plate there is a corre-
sponding normal effective force directed from
line AB normal to axis OZ, and each force is
proportional to the mass of its plate. As
shown in the last paragraph of Art. 43, the resultant of this
system of parallel forces acts through the center of gravity of the
body. Its amount is equal to 2 dMru2 = Mrco2, since f and co
are constants.
If the axis OZ coincides with the line of symmetry, r = 0 and
the resultant normal effective force Mrco2 = 0.
Case 3. // a slender prismatic rod of length I is rotating about
an axis through one end at any angle 6 with the axis, the resultant
normal effective force is equal to Mru>2 and acts through a point
distant f I from the point of support.
FIG. 249
166 APPLIED MECHANICS [CHAP, ix
In Fig. 250, let OB be the rod of length I with its center of
gravity at C, and let it be rotating about the axis OA with
angular velocity co. Let the rod be divided into
0 equal elementary parts. The normal effective
force on any elementary mass dm is equal to dMpa2,
p being the radial distance of the mass dM. These
forces are proportional to the radial distances of
the masses and are acting normal to the axis
AO. Since co2 is the same for all elements, the
resultant of all the effective forces becomes co2
dMp = Mru2 and acts at point Q, distant f I from 0. (See Art.
48.) If the axis does not pass through the end of the rod, the
part on each side of the axis is treated independently.
Problem 1. A steel cone has the following dimensions: height h = 4
inches; radius of base r = 2 inches. If it is rotating at 200 r.p.m. about an
axis parallel to its geometric axis and distant 6 inches from it, what is the
resultant normal effective force? Ans. 32.4 Ibs.
Problem 2. If the cone of Problem 1 is rotating about a diameter of the
base at 120 r.p.m., what is the resultant normal effective force?
Ans. 1.94 Ibs.
Problem 3. A cast iron disk 8 inches in diameter and 2 inches thick rotates
about an element of the cylindrical surface. Compute the normal effective
force for a speed of 60 r.p.m. Ans. 10.7 Ibs.
Problem 4. A steel rod 1 inch in diameter and 18 inches long is rotating
about a vertical axis through one end. If it stands at an angle of 45° with the
axis, at what speed is it rotating? Ans. 64.5 r.p.m.
107. Reactions of Supports of Rotating Bodies. For any
given problem in rotation, the unknown reactions of the supports
may be determined by the principle of Art. 85. That is, if in
addition to the actual impressed forces there be added the reversed
effective forces, the body will be under static conditions and all
static equations of equilibrium will be true. The method of solu-
tion is as follows: Draw the free body diagram and show all im-
pressed forces, known and unknown. (Unknown reactions of
supports are commonly replaced by axial, normal and tangential
components.) Determine the angular acceleration and velocity
from the given conditions. Determine the normal and tangential
components of the effective force and apply these reversed. These
/b2
components act through a point at a distance — from the axis.
ART. 107]
ROTATION
167
Solve as many equations of equilibrium as are necessary to deter-
mine all of the unknown quantities.
If r = 0, Mfco2 = 0 and Mfa = 0, so there are no kinetic re-
actions of the supports for Cases 1 and 2, but the reactions are
the same as when the body is at rest. In Case 3, if the axis
passes through the center of gravity of the rod the reaction of
the support becomes a couple.
EXAMPLE 1.
Fig. 251 (a) represents a steel disk 1 foot in diameter and 1 inch thick, free
to rotate about an element through 0. If it starts from rest with C vertically
above 0 and rotates under the influence of gravity alone, find the normal and
tangential components of the hinge reaction at 0 when 0 = 90°.
FIG. 251
Solution: — r = * ft.; W = 32.07 Ibs.; M = 0.995; Ic = \ Mr* = 0.124;
7o = 7C + Mr2 = 0.373.
The equation of motion is
Wr sin 6 = 70a.
When 6 = 90°, sin 6 = 1 and this equation becomes
32.07 X 0.5 = 0.373 a.
a = 43 rad. per sec. per sec.
To obtain co, solve for a in the equation of motion and substitute its value
in the expression add = to du and integrate between the proper limits.
Wr -WrcosO = i/"2.
When 6 = 90°, cos 6 =0 and this equation becomes
32.07 X 0.5 = 0.5 X 0.373 co2.
co = 9.27 rad. per sec.
Mra = 0.995 X 0.5 X 43 = 21.4 Ibs.
0.995 X 0.5 X 86 = 42.8 Ibs.
= -±- = 0.75 ft.
r Mr
distance OQ.
Fig. 251(b) shows the free body diagram in the position asked for, with the
effective forces reversed, and the hinge reaction represented by its normal and
tangential components. From the static equations of equilibrium,
SFX = 0, so Rn = 42.8 Ibs.
2FV = 0, so Rt = 32.07 - 21.4 = 10.67 Ibs.
168 APPLIED MECHANICS [CHAP, ix
EXAMPLE 2.
A vertical axle MN 4 feet long carries a horizontal arm AB 2 feet long which
is attached to the axle 1 foot from the top as shown in Fig. 252. On the end
of the horizontal arm is a cast iron sphere 6 inches
in diameter. The axle is rotated positively by a
pull of 10 Ibs. on the cord which passes around
the pulley C. If the sphere starts from rest in the
XZ plane and the pull is parallel to the X axis,
determine the reactions due to the sphere after one
revolution.
lOlbs Solution: — W = f irr3 X 450 = 29.45 Ibs.
M = 0.914. 1MN = 3.679. j = 2.013 ft.
0 = 2 TT radians = 360°.
^ 252 Equation Pd = I a gives
10X0.5 =3.679 a.
a = 1.36 rad. per sec. per sec.
Since the acceleration is constant,
co = V2 o0 = 4.135 rad. per sec. after 1 revolution.
Mrct = 0.914 X 2 X 1.36 = 2.486 Ibs.
Mrco2 = 0.914 X 2 X 17.1 = 31.25 Ibs.
The latter two forces are added reversed in Fig. 252, so the system as shown is
in equilibrium.
Equation 2FZ = 0 gives
Rz = W = 29.45 Ibs.
Equation 2MR = 0 gives
*#*' X 4 - 29.45 X 2 - 31.25 X 3 - 10 X 1 = 0.
Rz' = 40.66 Ibs.
Equation 2FX = 0 gives
Rx = 0.59 Ibs.
Equation 2MR = 0 gives
*Ry' X 4 - 2.486 X 3 = 0.
Ryr = 1.86 Ibs.
Equation 2Ftf = 0 gives
Ry = 0.62 Ibs.
Problem 1. With the same general data as in Example 1 above, compute
the normal and tangential components of the reaction when 0 = 45°.
Ans. Rn = 10.15 Ibs. Rt = 7.54 Ibs.
Problem 2. In Problem 1 compute the vertical and horizontal components
of the reaction when 0 = 135°. Ans. Ry = 73 Ibs. Rx = 62.3 Ibs.
Problem 3. In Problem 1 compute the normal and tangential components
of the reaction when 0 = 180°. Ans. Rn = 117.6 Ibs. Rt = 0.
Problem 4. With the same general data as in Example 2 above, compute
the reactions when the sphere has rotated one- half revolution from rest.
Ans. Rx = 18.33 Ibs. RJ = -23.93 Ibs. Ry= -0.62 Ibs. Ry'= -1.86 Ibs.
ART. 108] ROTATION 169
Problem 5. With the same general data as in Example 2 above, compute
the reactions when the sphere has rotated one second from rest.
Ans. 0 = 0.68 rad. Rx = -2.9 Ibs. R,' = 17.1 Ibs. Ry = 9.2 Ibs. Ry' =
-9.44 Ibs.
108. Compound Pendulum. Any physical body suspended
from a horizontal axis not passing through the center of gravity
and free to rotate under the influence of gravity and the reaction
of the support is called a compound pendulum. The ^v
student should refer again to Art. 95, Simple Circular 0)
Pendulum.
Let Fig. 253 represent a compound pendulum of
weight W, suspended at 0, and let (7 be its center
of gravity. Let I be its moment of inertia with re-
spect to the axis of rotation, k its radius of gyration,
r the distance from the support to the center of
gravity and a the angular acceleration. The equation of moments
about 0 gives
= -Wrsin6 = la.
W
Since / = Mk2 = — k2,
Wr sin 9 _ rg sin 6
~^7 ~*~
9
The tangential acceleration at of point Q, at a distance I from
the axis 0, is
rig
k2
If the length I be taken equal to — , the tangential acceleration a t
of point Q will be
at = —gsinO,
which is the same as the acceleration of the simple circular pen-
dulum. It is seen from this that a simple circular pendulum of
k2
length I = — will vibrate in the same time as the compound pen-
k2
dulum. The length — is called the length of the compound pen-
dulum, and the point Q is called the center of oscillation.
170 APPLIED MECHANICS [CHAP, ix
k2
Since I = — , the time of one complete period becomes
T = 9
g
rri
AlSO, k = jr- Vj/P.
J 7T
The point of suspension and the center of oscillation are inter-
changeable, as will now be shown. Let kc be the radius of gyra-
tion of the pendulum with respect to the axis through the center
of gravity C parallel to the axis of rotation. Since
k<2 = k2 _ p and ki = j^
k? = r (I - r) = OC X QC.
Since kc is a constant, regardless of the point of suspension, the
product OC X QC must be a constant. If OC is made smaller,
QC becomes proportionately larger, and vice versa.
Again, this equation would not be altered in any way if the
position of 0 and Q were interchanged, hence if the pendulum is
inverted and suspended from Q, point 0 must become the center
of oscillation. Since the length I remains the same, the time of
vibration is the same.
EXAMPLE.
A cylinder 2 inches in diameter and 12 inches long is hung from an axis
through the diameter of one end. Find the time of oscillation. From what
other point could it be suspended to vibrate in the same length of time?
Solution: — k2 = + * = 0.335 ft.2.
r-2-Vi^-o.«*~
The center of oscillation is given by
7 = 0.5 :
If the cylinder is suspended from the center of oscillation it will vibrate in
the same length of time.
Problem 1. Find the time of oscillation and the center of oscillation for
the cylinder of the Example above, if suspended from an axis through the
cylinder 1 inch above the middle. Arts. T = 1.16 sec. — = 1.105 ft.
Problem 2. A bar 1 inch square is suspended from a central axis through
one end parallel to the edges. What must be its length in order that it shall
oscillate in 1 second. (Complete period.) Ans. h = 1.225 ft.
Mrot .
reversed
ART. 110] ROTATION 171
Problem 3. Show that the bar of Problem 2 has four points of suspension
for which its period of oscillation is 1 second. Arts. 0.408 ft. from the end.
109. Center of Percussion. Let P, Fig. 254, be an impulsive
force or blow which causes angular acceleration of the body which
is suspended from 0. The force P is variable, but at any instant
during the blow,
Pd = la.
By the principle of Art. 105, the resultant tangential effective
k2
force Mra acts at point Q, distant — from the point of support.
Let Rx be the tangential component of the reaction at 0 caused by
P. If the effective force Mra is applied reversed, y
the system shown will be in equilibrium. By using fi* i+
the equation HMg = 0, the value of Rx may be
computed for any value and position of force P. p
It is evident that if force P is applied at point Q, ^"
Rx = 0. The point Q at which the body may be
struck without producing any reaction parallel to Li-
the tangent is called the center of percussion. It
is coincident with the center of oscillation of a pen-
dulum. It follows, then, that the center of percussion and the
point of suspension are interchangeable.
Problem 1. A slender rod 2 feet long suspended in a vertical position from
an axis through one end has a force of 100 Ibs. applied normal to the rod at its
middle point. What is the amount and direction of the horizontal kinetic
reaction caused by the force? Ans. Rx = 25 Ibs.
110. Centrifugal Tension in Flywheels. If the tension in the
arms of a flywheel is neglected, the tensile stress in the rim due
to rotation may be computed. For the half rim shown in Fig. 255
the normal effective force Mrco2 acts through the center of gravity.
Also, the effective force reversed as indicated would be in equilib-
rium with the two induced tensile forces, P, P. Then, if W is the
weight of the half rim,
If r is the mean radius of the rim, r = — (approximately). The
• 7T
p
unit stress s = -j, A being the area of the cross section of the rim.
A.
172
APPLIED MECHANICS
[CHAP, ix
If the tension in the rim of the flywheel is neglected, the tensile
stress in the arms may be computed. In Fig. 256, let AB be the
part of the rim carried by one arm. Let W be the weight of this
part and let r be the distance from 0 to its center of gravity.
The induced tensile force caused by rotation is
g
P
The unit stress is s = -r , A being the area of the cross section of
A.
the arm.
FIG. 255
FIG. 256
Problem 1. Show that if the tension in the arms is neglected, the speed
necessary to produce the same stress in the rims of flywheels varies inversely
as their radii.
Problem 2. A cast iron flywheel 12 feet in diameter has a rim 2 inches
thick and 12 inches wide. If the flywheel is rotating at 200 r.p.m., what is
the unit centrifugal tensile stress in the rim if the tension in the arms is neg-
lected? Ans. 1490 Ibs. per sq. in.
Problem 3. At what speed must a flywheel 1 foot in diameter be rotated
to produce the same unit stress in the rim as in Problem 2?
Ans. 2400 r.p.m.
111. Weighted Conical Pendulum Governor. Fig. 257 (a)
represents a weighted conical pendulum governor which consists of
two spheres, A, A, at the ends of arms BA, and a weight W\ sup-
ported by the collar CC. Consider first the weight W\ as the
free body, Fig. 257 (b), with the governor rotating uniformly.
The body is under static conditions so equation £Fy = 0 gives
W,
P =
2 cos 0i
Consider next one of the spheres and its arm as the free body,
Fig. 257 (c). The impressed forces acting upon the free body are
three in number, the weight W, the tension P and the pin reaction
at B. If the effective force Mru>2 is added to the system reversed
ART. 112]
ROTATION
173
in direction, the free body will be under static conditions. Accu-
rately, the force Mrco2 should act through the center of oscillation
of the sphere and its arm and the weight of the arm should be
considered, but the error is small if the weight of the rod be neg-
lected and Mrco2 be considered to act through the center of gravity
of the sphere. The equation SM# = 0 gives
Pd + Wr = Mrtfh.
Problem 1. In Fig. 257, let BD = 12 inches, DA = 4 inches, W = 20 Ibs.,
Wi = 100 Ibs. and 6 = 0i = 45° in the lowest position. At what speed will
the governor begin to act? Ans. 122 r.p.m.
Problem 2. If the governor described in Problem 1 has a spring which
carries half of the weight W\ when it is in its lowest position, at what speed
will it begin to act? Ans. 94.5 r.p.m.
112. Balancing of Rotating Bodies. It was seen in Art. 107
that if a body rotates about an axis not through its center of grav-
ity, the bearing reactions have kinetic components. These con-
tinually change in direction and so cause destructive vibration.
Balancing consists in adding rotating parts in such a way that
the effective forces for the entire system are in equilibrium and
no kinetic reactions are induced. The static reactions due to
gravity and any other constant impressed forces remain constant
whether the body is at rest or in motion.
In the following discussion, only rotation at constant speed will
be considered, and only rotation of bodies for which the resultant
174
APPLIED MECHANICS
[CHAP, ix
normal effective force passes through the center of gravity. (See
Art. 106.)
113. Balancing of Bodies in the Same Plane Normal to the
Axis of Rotation. Let A, Fig. 258, be a body of weight Wi, at
a radial distance r\t rotating about the axis through 0 normal to
OA with angular velocity co. By its rotation it exerts upon the
Wi
axis O a centrifugal pull equal to — nw2 which continually changes
y
in direction and causes variable reactions at the supports. If, how-
FIG. 258
FIG. 259
ever, another body of weight Wi be placed diametrically opposite
in the plane of rotation at the end of radius r2, of such length that
9 9
the two bodies are in balance since the centrifugal pulls are equal
and opposite. Since co and g are constants, the condition above
reduces to
TF2r2 = Wn.
It should be noted that the condition TF2r2 = W\TI is also the
condition for static balance. In practice it is customary to deter-
mine the necessary value of TF2r2 by means of static balancing.
Fig. 259(a) represents a number of bodies, TFi, TF2, TFs, at
radial distances n, r2, r3, in the same plane, rotating about the
axis through 0 normal to their plane. These are to be balanced
by a single weight W with radius r. If a vector polygon, Fig.
259(b), be drawn in which a, b and c represent W\TI, TF2r2 and
Wzn respectively in magnitude and direction, the closing line d
will represent Wr in magnitude and direction. Either W or r
may be assumed and the other computed.
Problem 1. A sphere weighing 40 Ibs. at a radial distance of 10 inches from
its axis of rotation is to be balanced by another of 60 Ibs. weight. What must
be its radial distance? Ans. 6| in.
ART. il4j ROTATION 175
Problem 2. In Fig. 259, TFi = 10 Ibs., W* = 5 Ibs., W3 = 20 Ibs., n = 24
inches, r2 = 20 inches, r3 = 16 inches, angle AOB = 45° and angle BOG = 90°.
If r is to be 12 inches, what must be the weight W to balance the system?
What is the angle DOA7 Ans. W = 25.7 Ibs. DO A = 105° 50'.
114. Balancing of Bodies in Different Normal Planes. In
either Fig. 260(a) or 260(b) let Wi be the body to be balanced,
and 00 the axis about which it rotates. Let planes through any
two points B and C normal to the axis be the planes in which the
balancing bodies are to lie. If W\ is to be balanced by two bodies
Wz and TF3, they must all lie in a plane containing the axis 00. It
is also necessary that the sum of the moments of the normal
effective forces about any point in this plane shall be equal to
zero.
FIG. 260
Equation 2MB = 0 gives
W* 2, W
— r3oj2o =
Q 9
or W^Tsb = W\r\a.
In Fig. 260(a), equation 2MC = 0 gives
W2r2b = F,r,(6 - a).
In Fig. 260(b), the same equation gives
W2r2b = Wiri(b + a).
From these equations the two unknown quantities, W^ and
TFsr3, may be determined.
Similarly, any other bodies Wi, W\ft etc., in any other planes
normal to the axis may be balanced by bodies W 2', TF2", etc., in
the plane through B, and bodies TF3', TF3", etc., in the plane through
C. Then finally all the bodies W 2, W 2', TF2", etc., in the normal
plane through B may be replaced by a single body, and the bodies
Wa, WJ, TF3", etc., in the normal plane through C may be replaced
by a single body.
176 APPLIED MECHANICS [CHAP, ix
Problem 1. A shaft 6 feet long between bearings carries a steel disk 2 feet
in diameter and 4 inches thick, 2 feet from the left bearing. The disk is keyed
to the shaft and is eccentric 4 inches. What weights must be added in planes
6 inches from the bearings, at a radial distance of 1 foot, in order to balance
the disk? Ans. 1 19.7 Ibs. at left. 51.3 Ibs. at right.
Problem 2. If on the shaft described in Problem 1 another similar disk is
placed 2 feet from the right end and 90° back of the first, what must be the
balancing weights in order to balance both disks? At what angle with the
position of the balancing weights in Problem 1 must they be placed?
Ans. Left, 130.1 Ibs., at 23° 10'. Right, 130.1 Ibs., at 66° 50'.
GENERAL PROBLEMS.
Problem 1. A pulley 2 feet in diameter rotating at 600 r.p.m. is brought to
rest in 50 seconds by a constant force of friction on its shaft. How many
revolutions does it make? Ans. 250.
Problem 2. What is the tangential acceleration of a point on the rim of the
pulley in Problem 1? What is the normal acceleration and the tangential
velocity at the end of 10 seconds?
Ans. at = 1.257 ft. per sec. per sec. an = 2528 ft. per sec. per sec. v =
50.3 ft. per sec.
Problem 3. The drum of a hoisting engine for a mine cage is 50 inches in
diameter. If the cage is to be lowered at the rate of 20 feet per second, how
many r.p.m. must the drum make? Ans. 91.7.
Problem 4. If the mine cage in Problem 3 weighs 500 Ibs. and the moment
of inertia of the drum is 120, during what time may the cage be allowed to drop
freely before the given velocity is obtained? Ans. 1.73 sec.
Problem 5. A flywheel weighing 200 Ibs. has its axis of rotation $ inch from
its geometric axis. If the wheel is midway between two bearings, what are
the kinetic reactions at the bearings when the flywheel is rotating at 600 r.p.m.?
Ans. 128 Ibs. on each.
Problem 6. If the axis of the flywheel in Problem 5 is horizontal, what
speed would be necessary to cause each reaction to vary from zero to 200 Ibs.
during a half revolution? Ans. 531 r.p.m.
Problem 7. A vertical shaft 6 feet long carries a weight of 100 Ibs. 4 inches
from its axis, 1 foot from the top support, and 50 Ibs. 6 inches from the axis,
4 feet from the top support on the opposite side of the shaft. Find the normal,
ROTATION 111
tangential and axial components of each reaction when the shaft is rotating at
80 r.p.m.
Ans. At bottom, Rn = 25.6 Ibs. Rt = 0. Rv = 150 Ibs. At top, Rn =
43.6 Ibs. Rt =0. Rv = 0.
Problem 8. If the weighted shaft described in Problem 7 is brought to
rest by a brake in 5 seconds, what are the tangential reactions?
Ans. Rt = 0.58 Ib. at bottom. Rt = 1.01 Ibs. at top.
Problem 9. A steel rod 1 inch in diameter and 10 feet long is supported
on a horizontal axis through one end normal to the axis of the rod. Locate the
center of oscillation. What is its period of oscillation if it is used as a pendulum
with small amplitude? Ans. 6.6668 ft. T = 2.86 sec.
Problem 10. If the rod described in Problem 9 is raised to the horizontal
position and then released, what are the normal and tangential reactions at
the instant of release? What are the normal and tangential reactions as it
passes the vertical position?
Ans. Rn = 0. Rt = 6.675 Ibs. Rn = 66.8 Ibs. Rt = 0.
Problem 11. If the rod described in Problem 9 is released from the vertical
position above the axis, what are the normal and tangential components of
the reactions when it is 45° from the vertical? When in the horizontal posi-
tion? When in the lower vertical position?
Ans. At 45°, Rn = 7.11 Ibs. Rt = 4.74 Ibs. At 90°, Rn = 40.1 Ibs.
Rt = 6.675 Ibs. At 180°, Rn = 106.8 Ibs. Rt = 0.
Problem 12. A cast iron flywheel rim weighs 4000 Ibs. and is cast in two
parts. The two sections are to be held together by six steel bolts at each
joint. If the mean radius of the rim is 5 feet and the maximum speed is to be
240 r.p.m., compute the necessary area of each bolt at the root of the thread
for an allowable unit tensile stress of 12,000 Ibs. per sq. in.
Ans. 0.87 sq. in.
Problem 13. A solid cast iron flywheel rim is 4 inches wide, 3 inches deep
and 24 inches outside diameter. If the tension in the arms is neglected, what
is the unit tensile stress in the rim when the wheel is rotating at 1600 r.p.m.?
Ans. 2080 Ibs. per sq. in.
Problem 14. A cast iron flywheel 10 feet in diameter has a rim 10 inches
wide and 6 inches deep. If the ultimate tensile strength of cast iron is 25,000
Ibs. per sq. in., what speed will rupture the wheel if the tension in the arms is
neglected? Ans. 1020 r.p.m.
Problem 15. The flywheel described in Problem 14 has a total weight of
8000 Ibs. If the center of gravity is 1 inch from the axis of rotation and the
wheel is midway between two supports, what is the variation in each reaction
when the wheel is rotating at 60 r.p.m.? Ans. 817 Ibs.
Problem 16. The wheel of the Brennan monorail car gyroscope weighed
1000 Ibs. and was rotated at 3000 r.p.m. on a horizontal shaft midway between
bearings. If it had been eccentric 0.01 inch, what would have been the varia-
tion in each reaction?
Ans. From 1777 Ibs. upward to 777 Ibs. downward.
Problem 17. A 20-lb. governor ball on an arm 3 ft. long rotates at such a
speed that the arm is kept at an angle of 45° with the axis. What is its speed?
What is the tension in the arm? Ans. 37.2 r.p.m. 28.3 Ibs.
178
APPLIED MECHANICS
[CHAP, ix
Problem 18. If the length of the arm of the governor ball described in
Problem 17 is reduced to 2.5 feet and the governor is rotated at the same speed,
what effect is produced on the angle with the axis and on the tension in the
arm? Ans. 6 = 32°. T = 23.5 Ibs.
Problem 19. Fig. 261 represents a weighted conical pendulum governor
for which Wi = 40 Ibs. and W — 10 Ibs. What speed will keep the governor
in the position shown? What is the tension in each arm?
Ans. 158 r.p.m. 30.2 Ibs. in lower. 45.2 Ibs. in upper.
FIG. 261
FIG. 262
Problem 20. If 6 = 30° when the governor shown in Fig. 261 is in its low
position, at what speed will it begin to act? Ans. 130 r.p.m.
Problem 21. In xthe swing shown in Fig. 262 each car weighs 1200 Ibs. As
the cars are rotated about the vertical axis AB they swing out from the vertical.
If the maximum allowable value of 6 is 45°, what is the maximum speed at
which it may be run? What is the corresponding stress in the supporting
cables? Ans. 13.5 r.p.m. 1700 Ibs.
W/
-5- -3'
(a)
FIG. 263
Cb)
Problem 22. Fig. 263 (a) shows a side view and Fig. 263 (b) an end view of
a shaft to which two weights Wi and Wi' are attached. If Wi = 100 Ibs.,
TI = 18 inches, Wi = 50 Ibs. and r/ = 15 inches, what weights at A and B
with radii of 12 inches will be necessary to balance the system? What is the
angle of each radius with the horizontal plane?
Ans. WA = 170 Ibs. BA = 262° 5'. W B = 43.1 Ibs. 8B = 154° 10'.
CHAPTER X.
COMBINED TRANSLATION AND ROTATION.
115. Any Plane Motion Equivalent to Combined Translation
and Rotation. In a plane motion of a body, each point of the
body remains at a constant distance from a fixed plane. This
plane or any parallel plane may be called the plane of motion of the
body. The plane of motion through the center of gravity of the
body is commonly used for reference.
Any plane displacement of a body may be considered to be made
up of a rotation about any point in the plane of motion and a
corresponding translation. That is, the same result would have
been obtained by the two simple motions as by the actual motion,
whatever it may have been. In Fig. 264 let AB be a line in the
plane of motion joining any two points of a body in their original
position and let A2B2 be their position after any plane motion of
the body. Let 0 be any point in the plane of motion. The
displacement from AB to A2B2 may evidently be made by a
rotation about 0 to the position AiBi parallel to A2B2, then a
translation from AiBi to A2B2.
FIG. 264
Again, any plane displacement of a body is equivalent to a simple
rotation about some fixed point in space. In order to locate this
point, join AA2 and BB2, Fig. 265. Erect perpendicular bisectors
of AA2 and BB2 which intersect at 0. The triangles AOB and
A2OB2 are equal in all their parts. Therefore angle AOA2 =
angle BOB2) since angle AOB = angle A2OB2. Hence it is plain
that the displacement from AB to A2B2 is equivalent to simple
rotation through angle AOA2 about point 0.
179
180
APPLIED MECHANICS
[CHAP, x
If AAZ and BB2 are parallel, the point 0 is at infinity and the
motion is equivalent to pure translation.
Problem 1. In Fig. 264, resolve the displacement from AB to A2B2 into
a rotation about 0' and a corresponding translation. Do the same, using
point A as a center. Do the same, using the point midway between A and B
as a center.
Problem 2. A wheel rolls along a horizontal plane through one-fourth of
a revolution. Find the center of equivalent rotation. Do the same for
rotation through one-half of a revolution.
116. Resolution of Velocities in any Plane Motion. The
absolute velocity of any point of a body which has any plane
motion can usually be determined best by getting first its velocity
relative to some special point of reference on the body and then
the absolute velocity of the point of reference. By Art. 90 the
absolute velocity of the given point is equal to the vector sum of
the two velocities, its own velocity relative to some point of refer-
ence, and the absolute velocity of the point of reference.
Let B, Fig. 266, be any point of a rigid body and let A be the
point of reference chosen. Let Vi be the absolute velocity of A
and vr the velocity of B relative to A. Since A and B are fixed
points on the rigid body, the only velocity B can have relative
to A is tangential. This is equal to rco, r being the length AB
and co the angular velocity. By the principle stated above, the
absolute velocity of B is the vector sum of the two, or v.
V, '• rev'
FIG. 267
Conversely, the absolute velocity of any point of a rigid body
may be resolved into two components, one of which is equal and
parallel to the absolute velocity of any chosen point of reference
on the body, while the other is normal to the line joining the
two points.
As an example consider the wheel shown in Fig. 267 which is
rolling to the left on a horizontal plane. If the velocity of the
ART. 117] TRANSLATION AND ROTATION 181
center of the wheel is v\ and its angular velocity is co, the velocity
of any point on the rim with respect to the center is no = Vi for
free rolling. The absolute velocity of any point is the vector sum
of its velocity with respect to the center and the absolute velocity
of the center, as shown at B. The absolute velocity of B is v, the
vector sum of Vi and rco. In the same way the absolute velocity of
the bottom point C is the vector sum of its velocity with respect
to 0, which is rco, and the absolute velocity of 0, which is v\.
Since these are equal in amount and opposite in direction, the
absolute velocity of point C is zero.
In discussing the motion of the rolling wheel, it is sometimes
simpler to use point C as the point of reference. Since its absolute
velocity is zero, the absolute velocity of any point, as B, is the
same as its velocity relative to point C. Then VB — rBu, rB being
the distance BC.
Problem 1. In Fig. 266, Vi is 20 feet per second horizontal to the right,
r is 3 feet, w is 5 radians per second clockwise and AB makes an angle of 30°
with the horizontal. What is the absolute velocity of £?
Ans. 30.4 ft. per sec. at 25° 20' with hor.
Problem 2. A cylinder 1 foot in diameter is rolling to the right on a hori-
zontal plane with a uniform velocity of 10 feet per second. Using the center
as the point of reference, find the absolute velocity of a point on the rim in
front, 45° above the horizontal through the center. Check by using the
bottom point of the cylinder as the point of reference.
Ans. 18.46 ft. per sec. at 22° 30' with hor.
117. Resolution of Accelerations in any Plane Motion. The
principle of Art. 90, which was referred to in the preceding article,
is true for accelerations as well as for velocities. That is, the
absolute acceleration of any given point of a body is equal to the
vector sum of the relative acceleration of the given point with
respect to some chosen point of reference on the body and the
absolute acceleration of the point of reference.
Let B, Fig. 268, be any point of a rigid body and let A be the
point of reference whose absolute acceleration is a\. Let AB = r
and let the angular velocity and acceleration be co and a respec-
tively. The relative acceleration is most easily determined by
means of its tangential and normal components. The tangential
component is ra and the normal component is ro>2. These two
combined give the relative acceleration a', and finally a' and a\
combined give vector BC = a, the absolute acceleration of B.
182 APPLIED MECHANICS [CHAP.X
Conversely, the absolute acceleration of any point may be
resolved into three components, one of which is equal and parallel
to the absolute acceleration of any chosen point of reference on
the body, another equal to rco2 along the line joining the two
points, and a third equal to ra perpendicular to this line.
FIG. 268 FIG. 269
As an example consider the wheel shown in Fig. 269, which is
rolling to the left on a horizontal plane. Let the acceleration of
the center of the wheel be a\, its angular acceleration a and its
angular velocity co. The tangential component of the acceleration
of any point on the rim, as 5, relative to the center is ra = a\ for
free rolling. The normal component of the relative acceleration
is rco2. These two vectors combined give a', the relative accelera-
tion of B with respect to the center. The vector sum of a' and ai
gives a, the absolute acceleration of point B.
Problem 1. In Fig. 268, ai is 50 feet per second per second horizontal
to the right, AB is 3 feet long and is at an angle of 30° with the horizontal,
co is 5 radians per second clockwise and a is 20 radians per second per second
clockwise. Determine the absolute acceleration of point B.
Ans. a = 145.6 ft. per sec. per sec., 5° 45' above hor.
Problem 2. Solve for the absolute acceleration of the point described in
Problem 2 of Art. 116, using the same two points of reference.
Ans. a = 200 ft. per sec. per sec. toward the center.
118. Instantaneous Axis. Let A and B, Fig. 270, be any two
points of a rigid body having any plane motion. Let their veloci-
ties be in the directions of VA and VB as shown.
At A erect AO perpendicular to VA, and at
" VB B erect BO perpendicular to VB. Since the
absolute velocity of point A is normal to the
F 270 ' *s ecluivalent to rotation about
some point on AO. Similarly, since the
absolute velocity of point B is normal to BO, it is equivalent
to rotation about some point on BO. Since this point is on both
ART. 119] TRANSLATION AND ROTATION 183
AO and BO, it must be at their point of intersection 0. For the
instant considered it is therefore the center of rotation of points
A and B, and likewise of the rigid body upon which they are
located. It is called the instantaneous center of the body and the
axis through this point, normal to the plane of motion, is called the
instantaneous axis of the body.
In general, the instantaneous center will be at a different point
the following instant. Its path relative to the body is called the
body centrode and its path in space is called the space centrode.
In instantaneous rotation, as |n_ simple rotation, the relation
v = no holds true. Then VB = OBu and VA = OAco, co being the
angular velocity of the body at that instant.
Problem 1. AB in Fig. 270 is a uniform rod at an angle of 30° with the
horizontal. The vector VB is horizontal to the right and the vector VA is
vertical, upward. Locate the instantaneous center. If VB is 100 feet per
second and AB is 6 feet, what is tu? Ans. VA = 173.2 ft. per sec.
Problem 2. In Problem 1, what is the normal acceleration of the center of
gravity of the rod? Ans. -an = 3333 ft. per sec.
Problem 3. Take two pieces of cardboard about 6 inches square and
place one on top of the other with their edges coinciding. Let the bottom one
remain stationary while the upper left-hand corner of the top one is moved
downward along the left edge of the bottom one, and the lower left-hand corner
of the top one is moved to the right along the lower edge of the bottom one.
Locate on the upper card the instantaneous center for a number of positions
and prick through into the lower card. Draw the body and space centrodes
and cut the cards along these lines. Note that a straight line motion of the
corners is obtained by rolling the body centrode upon the space centrode.
119. Equations of Motion. As shown in Art. 117, the abso-
lute acceleration of any point of a rigid body having any plane
motion may be resolved into three components, one equal and
parallel to the acceleration of any
chosen point of reference on the body,
another equal to rco2 acting along the
line joining the two points, and a
third equal to ra acting perpendicular
to this line. If the mass of any par-
ticle is dM, the effective force for it is <
equal to the resultant of the three
components, dMa, dMru2 and dMra.
In Fig. 271 let A be the point of
reference and let the X axis coincide with a\y the absolute accel-
eration of A. Let 2F be the resultant of all the external forces,
184 APPLIED MECHANICS [CHAP, x
and SFj. and 2Fy the components of 2 F in the X and F directions
respectively. Since the external force system is equivalent to the
effective force system for the whole body of mass M,
%FX = I dMci! + I dMra sin 0 + / dMrco2 cos 0
= ai I dM + a fdMy + o>2 / dMx
= Mai + May + Mu?x.
2Fy = I dMra cos 0 - I dMrrf sin 0
= a I dMx -co2 I dMy
= Max — Mco2?/.
2MA = CdMr*a + idM^r sin 0
= a f r2 dM + ai f
If the center of gravity is taken as the point of reference, as is
usually the case, the three equations above become
SF, = Ma.
SFy = 0.
2 Mo = I0a.
70 is the moment of inertia of the body with respect to the axis
through the center of gravity.
The principles derived above may be stated as follows: —
1. In any plane motion the center of gravity is accelerated the
same as if the whole mass were concentrated at that point and acted
upon by forces equal in amount and direction to the actual forces.
2. In any plane motion the angular acceleration about the center
of gravity is the same as if that center were fixed and a couple of
moment = 2M0 applied to the body.
If the reversed effective forces are considered to be added to
the free body with its actual impressed forces, the equations of
equilibrium hold true. That is, if a force Ma be applied at the
center of gravity in a direction opposite to its absolute acceleration
and a couple IQa opposed in direction to the angular acceleration,
the problem is reduced to static conditions.
ART. 120]
TRANSLATION AND ROTATION
185
120. Wheel Rolling on Horizontal Plane. Several illustrative
examples of the last five articles will now be given. One of the
simplest is that of a wheel rolling freely on a plane surface. The
motion of the wheel is a combined rotation and translation.
If a horizontal force P, Fig. 272, acts at the center of the wheel
to produce perfect rolling on a rough surface, a frictional force F
is induced at the point of support. Then, by Art. 119,
P - F = — a.
Q
= Fr = Ia.
ZFX
Also
For perfect rolling, a = ra.
From these three equations the acceleration and the frictional
force may be determined. In the direction normal to the plane,
static conditions hold true, so W = N.
FIG. 272
H
FIG. 273
EXAMPLE 1.
In Fig. 272, let the wheel be a cylinder 4 feet in diameter, let W = 1000
pounds and P = 100 pounds. Determine F, a and a.
Solution: — By summing horizontal forces
1000
~ F =
By moments about 0,
1 1000
Also, a = 2 a.
By solution of these three equations,
F = 33| Ibs.
a = 2.147 ft. per sec. per sec.
a. — 1.073 rad. per sec. per sec.
EXAMPLE 2.
Fig. 273 represents a cylinder 2 feet in diameter weighing 200 pounds
resting on a horizontal plane surface. Attached to the cylinder and concentric
with it is a hollow cylinder 1 foot in diameter, of negligible weight, around
which a cord is wrapped. What horizontal force P applied to the cord as
186
APPLIED MECHANICS
[CHAP.X
shown will produce an acceleration of 20 feet per second per second? What
is the frictional force Ft Assume free rolling.
Solution: — Equation 1,FX = Max gives
Equation
= ha gives
P = 124.1 Ibs.
F = 0.
If P is applied lower than the point indicated, a frictional force is induced
in the direction opposite to P, while if it is applied above this point a frictional
force is induced in the same direction.
Problem 1. Solve Example 2 above if the small cylinder is solid and
weighs 50 pounds. Am. P = 147.5 Ibs. F = 7.7 Ibs.
Problem 2. A cast iron cylinder 1 foot in diameter and 1 foot long, free to
roll on a horizontal plane, has a force of 5 pounds acting horizontally at the
center normal to the geometric axis. Find the velocity and acceleration of a
point on the rim behind, 30° above the horizontal through the center, 5 seconds
from rest. What is the frictional force F?
Ans. v = 2.63 ft. per sec. forward and upward, 30° with hor. a = 4.87
ft. per sec. per sec. forward and downward, 24° 40' with hor. F = 1.67 Ibs.
Problem 3. Solve Problem 2 if the 5-lb. force is acting horizontally at the
top.
Ans. v = 5.26 ft. per sec. forward and upward, 30° with hor. a — 18.88
ft. per sec. per sec. forward and downward, 27° 15' with hor. F = 1.67 Ibs.
Problem 4. Solve Example 2 above if the diameter of the small cylinder is
1.5 feet. Ans. P = 106.5 Ibs. F = 17.7 Ibs.
121. Wheel Rolling on Inclined Plane. Let Fig. 274 repre-
sent a wheel rolling on a plane inclined at an angle ft with the
horizontal. The plane is considered to be
rough enough so that free rolling takes place.
If the wheel is released from rest at the top
of the plane and allowed to roll down freely
under the influence of gravity, there will be
acting in addition to W, the normal reac-
tion N and the frictional force F. The three
equations of motion may be written as in the
preceding article in order to determine F,
Let the X axis be parallel to the plane. Then
Fx = Wsmft-F = — a.
g
Fy = W cos ft - N = 0.
Fr = loot.
FIG. 274
N and the motion.
ART. 122] TRANSLATION AND ROTATION 187
Also, a = TO. for free rolling.
1 W
If the wheel is a cylinder, 70 = 5 — f2, so by solving for F,
« 0
F« JWsin/J.
If/ is the coefficient of static friction between the cylinder and the
plane, and F' = fN is the limiting value of the friction F,
, F' TF sin5 1
This is the limiting value for free rolling. If in any case J tan /3
is greater than the coefficient of friction /, slipping will take place.
If the cylinder slips, the friction is kinetic and a does not equal ra.
However, the force of friction becomes known, being fN (f is the
coefficient of kinetic friction), so all values may be determined.
Problem 1. A steel disk 3 inches in diameter and 1 inch thick is released
from rest on a 30° plane. If the coefficient of static friction is 0.25 and the
coefficient of kinetic friction is 0.20, determine its linear and angular velocity
when the disk has rolled 8 feet down the plane. What is the frictional force?
Ans. v = 13.1 ft. per sec. co = 104.8 rad. per sec. F = 0.334 Ib.
Problem 2. Determine the motion of the disk in Problem 1 if the plane and
disk are both smooth. Ans. v = 16.1 ft. per sec. co = 0. F = 0.
Problem 3. Solve Problem 1 if the angle of the plane is changed to 45°.
Ans. v = 17.08 ft. per sec. co = 68.2 rad. per sec. F = 0.283 Ib.
Problem 4. The steel disk described in Problem 1 is rolled up a 10° plane
by means of a horizontal force P applied at the center. If there is no slipping
and the acceleration is 2 feet per second per second, find the value of the force
P, the normal pressure N and the friction F. If at the end of 5 seconds the
force P is removed, how much farther up will the disk roll?
Ans. P = 0.544 Ib. N = 2.065 Ibs. F = 0.062 Ib. 13.4 ft.
Problem 6. A steel sphere 2 inches in diameter starts from rest and rolls
down a 15° plane. If the static coefficient of friction is 0.15 and the kinetic
coefficient of friction is 0.10, what is the linear and angular velocity after one
revolution? What is the force of friction?
Ans. v = 2.49 ft. per sec. co = 29.9 rad. per sec. F = 0.088 Ib.
122. Connecting Rod of Engine. Graphic Solution. The
principle of relative velocities and accelerations is especially well
adapted to the solution of the problem of the connecting rod of a
steam engine. In Fig. 275, A is the crosshead with velocity v
and acceleration a, these being the same as the velocity and
acceleration of the piston; AB is the connecting rod, of length I', B
is the crank pin with tangential velocity v\\ BO is the crank of
length r; 0 is the center of rotation of the flywheel. The fly-
wheel is assumed to be heavy enough so that point B has a rotation
188 APPLIED MECHANICS [CHAP.X
practically uniform, with angular velocity coi. The angle <f>
between the connecting rod and the line AO increases to a maxi-
mum when B is at the top point in its circle, then decreases and
changes to negative values.
FIG. 275
The motion of the connecting rod is an oscillatory rotation with
variable angular velocity co about point A, which meanwhile has
an oscillatory translation along the line AO. The absolute veloc-
ity of point B is vi = rcoi normal to OB. By the principle of
relative velocities, this is equal to the vector sum of the absolute
velocity of A and the relative velocity of B with respect to A.
Let vz be the velocity of B relative to A. It is known that its
direction is normal to AB, since A and B are rigidly connected,
and that the direction of v, the velocity of point A, is horizontal.
The vector diagram, Fig. 276, completely determines the value
of v% and wj. Since the motion of B relative to A is a rotation with
radius I, vz = Zco, or co = y . The linear velocity of A and the
i
angular velocity of the rod with respect to A are thus completely
determined, so the absolute velocity of any point on the rod may
be found.
vAbs.ofA
a
FIG. 276 FIG. 277
In Fig. 277 the unknown accelerations are determined. Since
point B, Fig. 275, is moving in a circle with uniform angular
velocity coi, its only acceleration is rcoi2, toward the center 0. This
is drawn to scale in Fig. 277. The relative acceleration of B with
respect to A and the absolute acceleration of A must have rcoi2 as
ART. 123] TRANSLATION AND ROTATION 189
their vector sum. The relative acceleration of B with respect
to A is made up of two components, one wholly known, the other
known only in direction. The normal component ko2 is com-
pletely known, so is drawn first. The tangential component la
is perpendicular to the direction of the connecting rod, and the
absolute acceleration a of point A is horizontal and must close
the polygon. This completely determines a and a, so the abso-
lute acceleration of any point on the rod may be found.
Problem 1. The crank of an engine is 1 foot long and the connecting rod
is 6 feet long. If wi = 20 radians per second, find the velocity and the accelera-
tion of the crosshead when 0=0°; when 6 = 30°.
Ans. v = 0. a = 467 ft. per sec. per sec.
v = 11.46 ft. per sec. a = 380 ft. per sec. per sec.
Problem 2. Solve for the velocity and acceleration of the crosshead of
Problem 1 when 6 = 90°; when 6 = 180°.
Ans. v = 20 ft. per sec. a = — 68 ft. per sec. per sec.
v = o. a = — 333 1 ft. per sec. per sec.
123. Kinetic Reactions on Connecting Rod. In order to
determine the crank pin and crosshead pin pressures, the connect-
ing rod is considered as a free body, Fig. 278. The impressed
forces acting upon the rod consist of the following: its weight W
vertically downward at its center of gravity; the pressure F
from the piston rod through the crosshead; the normal pressure
NA from the crosshead; the crank pin reaction at B. The crank
pin reaction is resolved into its two components, N along the rod
and T perpendicular to the rod. Of these impressed forces,
N, T and NA are unknown.
C- 7V\"r* ft* \
FIG. 278
Under the action of these impressed forces the rod is accelerated
both in translation and rotation. By the method of Art. 122 the
values of o>, a and a may be determined. If now the reversed
effective forces be added to the free body, Fig. 278, it will be
under static conditions as discussed in Art. 119. If coi is clock-
wise and the value of 6 between 0° and 90°, o>, a, v and a will be
in the directions shown. The three components of the accelerar-
190
APPLIED MECHANICS
[CHAP, x
tion of the center of gravity are a, horizontal to the right, ra
downward, normal to the rod, and fco2 along the rod toward A.
Then the reversed effective forces are (1) Ma horizontal to the
left; (2) Mrco2 outward away from A', and (3) Mr a upward,
normal to the rod. It will be remembered that Mr a acts through
fc2
the point Q, distant — from A and not through the center of
gravity. The force Mrco2 acts through the center of gravity as
shown in Case 1, Art. 106. Since all of the forces are known
except NA, N and T, these may be determined by the solution of
the three equations of equilibrium.
In Fig. 278, equation 2FX = 0 gives
F - N cos $ + T sin 0 - Ma + Mrco2 cos <£ - My a = 0.
Equation %FV = 0 gives
NA-W + Mxa + Mrco2 sin 0 - T cos <f> - N sin 0 = 0.
Equation 2M^ = 0 gives
Wx - May - MTrcL =\ + Tl = 0.
EXAMPLE.
In Fig. 278, let r = 1 foot, I = 6 feet, W = 200 pounds, r = 3.8 feet,
F = 10,000 pounds, o>i = 30 radians per second, 6 = 30° and IA = 120.
Solve for NA, N and T.
W 200
Solution: —
120
5.08 ft.
r Mr 6.21 X 3.8
= 0.0833; cos0 = 0.9965; 0 = 4° 47'.
x = 3.786 ft.; y = 0.316 ft.
Scale
^^ I "=600
t&h
'/sec*
FIG. 279
FIG. 280
The vector diagram for the velocities is shown in Fig. 279, from which by scale
v = 17.2 ft. per sec.,
/co = 26.1 ft. per sec.,
26.1
6
= 4.35 rad. per sec.
ART. 124] TRANSLATION AND ROTATION 191
In order to determine the accelerations, their vector diagram is drawn,
Fig. 280.
ro>i2 = 900; 1<J = 6 X 4.352 = 113.
Vector la scales 442, from which a. = 73.7 rad. per sec. per sec.
Vector a scales 855 ft. per sec. per sec.
Ma = 6.21 X 855 = 5310 Ibs.
MTCL = 6.21 X 3.8 X 73.7 = 1740 Ibs.
Mrco2 = 6.21 X 3.8 X 4.352 = 445 Ibs.
These are the three components of the effective force for the body, and if
applied reversed, as shown in Fig. 278, are in equilibrium with the impressed
forces.
Equation ~ZFX = 0 gives
10,000 - 5310 - 1740 X 0.0833 + 445 X 0.9965 + 0.0833 T - 0.9965 AT = 0.
Equation ~ZFy = 0 gives
NA - 200 -f 1740 X 0.9965 + 445 X 0.0833 - 0.9965 T - 0.0833 N = 0.
Equation *2MA = 0 gives
5310 X 0.316 - 200 X 3.786 + 1740 X 5.08 - 6 T = 0.
Solution of the last equation gives
T = 1627 Ibs.
This value substituted in the first equation gives
N = 5150 Ibs.
These two values substituted in the second equation give
NA = 477 Ibs.
The resultant pressure of the crank pin is given by
VAT2 + T2 = 5400 Ibs.
Problem 1. The connecting rod described in Problem 2, Art. 79, is 6 feet
long, the crank is 1 foot long, the horizontal pressure from the crosshead pin
is 6000 pounds and the engine is running at 180 r.p.m. Find the pressure of
the guide on the crosshead and the total pressure on the crank pin when
d = 45°. Ans. NA = 414 Ibs. Crank pin pressure = 4190 Ibs.
124. Kinetic Reactions on Side Rod. In the side or parallel
rod of a locomotive, each particle of mass dM, as at D, Fig. 281,
FIG. 281
has a motion of rotation about its own center, point C on the
line AAi, and a translation the same as point C. If the linear
velocity of the locomotive is constant, the absolute acceleration
192 APPLIED MECHANICS [CHAP.X
of point D is ro>2 directed toward point C, co being the angular
velocity of the wheels. Since at any instant the accelerations ot
all the particles of the rod are the same in amount and direction,
the resultant of all the elementary effective forces is equal to their
sum, Mrco2, and acts through their center of gravity.
Upon the rod as a free body, Fig. 282, the impressed forces acting
are the two crank pin pressures and its weight W. The effective
force Mru>2 if added to the impressed forces reversed in direction
will give static conditions. The crank pin pressures are most
easily determined in terms of their two components, one vertically
W
upward equal to -^ , the other radial equal to — ~ — . It is evident
that the resultant reaction is a maximum when the rod is at the
bottom of its travel. At this point,
W
The reaction is a minimum when the rod is at the top, where
_ Mro?2 W
~ 2 — ~ ~2~*
Problem 1. A side rod weighs 420 pounds, the drive wheels are 6 feet in
diameter, the length of the crank is 16 inches and the locomotive is running at
a speed of 70 miles per hour. What is the maximum pressure on each crank
pin? Ans. 10,390 Ibs.
125. Kinetic Reaction on Unbalanced Wheel. If a wheel
whose center of gravity does not coincide with its geometric center
rolls along a horizontal plane surface, the re-
action of the surface is not constant in amount,
but changes during each revolution from a
value greater than the weight W to one less
than W. Let the wheel be as shown in Fig.
283 with its center at 0 and its center of
gravity at C, due to the added weight on one
side. Let the speed of the center be constant
FIG 28S
toward the left and let its amount be v. Since
the pull of gravity tends to retard the motion for values of 8 from
0° to 180° and to accelerate it for values of 8 from 180° to 360°,
the forces FI and F are necessarily variable.
By the principle of relative motion, the absolute acceleration of
point C is equal to the vector sum of the absolute acceleration of
ART. 126]
TRANSLATION AND ROTATION
193
the center, ra, and the relative acceleration of C with respect to
the center. The two components of the relative acceleration are
fa and rco2. Since a is zero, the acceleration of point C is no2
toward the center 0. If now the reversed effective force Mrco2 be
added, the wheel will be under static conditions and the equations
of equilibrium may be written. Equation 21 Fy = 0 gives
N = W + Mrco2 cos 0.
When 6 is zero the value of AT" is a maximum and when 8 is 180° it
is a minimum.
Problem 1. In Fig. 283, let r = 3 feet and r = 6 inches. At what speed
will the reaction N be zero when C is directly above 0?
Ans. 24.06 ft. per sec.
Problem 2. A locomotive drive wheel 6 feet in diameter weighs 2000
pounds and carries an axle load of 13,000 pounds. When the side rod and the
connecting rod are removed the center of gravity of the wheel is 0.6 of a foot
from the center of the wheel due to the counterweight. If the locomotive is
pulled by another at a speed of 30 miles per hour, what is the variation in the
pressure on the track during one revolution?
Ans. 23,000 Ibs. max. to 7000 Ibs. min.
126. Balancing Reciprocating Parts. A simple illustration of
the balancing of reciprocating parts is furnished by the slotted
FIG. 284
FIG. 285
slider apparatus driven by a crank rotating at constant speed, as
W
shown in Fig. 284. Let W be the weight and M = — be the
y
mass of the slider and let friction be neglected. If the angular
velocity of the crank AO is coi, the acceleration of the crank pin
is rcoi2 toward the center. The slider has a variable horizontal
acceleration a = rui2 cos 6, and its motion is simple harmonic.
The force to cause this acceleration is the variable pressure of the
crank pin at A and is equal to
p = M a = Mrcoi2 cos 0,
as shown in Fig. 285 (a). For values of 0 between 90° and 270°
the acceleration is toward the left, so the force P is acting toward
194 APPLIED MECHANICS [CHAP, x
the left on the slider. The equal and opposite pressure of the
slider on the crank pin is P', as shown in Fig. 285 (b), which is in
turn transmitted to the support at 0.
In order to balance the force P' on the crank pin, a mass M\ at
a distance r\ may be added opposite to the crank, as shown in
Fig. 286. If the values of MI and n are such that
MiTi = Mr,
the force P' is completely balanced, since the mass MI is exerting
a centrifugal force equal to Miricoi2 in the direction OMi and this
has a horizontal component of Miricoi2 cos 6.
The vertical component Mir^i2 sin 0 of the centrifugal force of
MI is not balanced, however. This force is a maximum at the
top and bottom points, where it is
equal to Miricoi2. If the value of Mxri
were half that of Mr, one-half of the
horizontal force P' would be balanced
and the vertical force at the top and
bottom positions would be only
\ Miricoi2. Since generally the hori-
zontal force is more injurious to the
mechanism than the vertical, the usual
practice is to balance about two-thirds of the horizontal force.
This leaves an unbalanced horizontal force of J Mrui2, and gives
an unbalanced vertical force of f Miricoi2.
In the ordinary reciprocating engine with connecting rod the
acceleration of the piston at the head end of the cylinder is greater
v^ r
than rcoi2 by the amount Zco2 = lj-2 = -= (rcoi2), and at the crank
i L
end is less than rcoi2 by the same amount, as may be shown by
the method of Art. 122. The mean value is rcoi2, so the recipro-
cating parts of such an engine are balanced in the same manner
as those of the slotted slider.
Problem 1. The total weight of a single-cylinder horizontal engine is
20,000 pounds and the weight of the reciprocating parts is 1000 pounds. The
crank is 1 foot long and the engine is running at 300 r.p.m. If the engine is
perfectly free to move, what will be the approximate amplitude of oscillation
of the frame and the magnitude of the displacing force at the end of the stroke?
Assume that the reciprocating parts have simple harmonic motion.
Ans. 0.63 inch. 30,650 Ibs.
Problem 2. If the engine of Problem 1 is balanced according to general
practice, what will be the maximum vertical unbalanced force when running
at 300 r.p.m.? Ans. 20,430 Ibs.
ART. 128]
TRANSLATION AND ROTATION
195
127. Balancing Both Rotating and Reciprocating Parts. In
the usual type of engine with connecting rod, the rod has a com-
bined rotation and translation. For the purposes of balancing,
the small crosshead end of the rod and one-half of the plain part
of the rod are considered to have a motion of translation with the
crosshead, crosshead pin, piston rod and piston. The large crank
end of the rod and the remaining half of the plain part of the rod
are considered to have a motion of rotation with the crank and
crank pin. In ordinary steam-engine construction the former is
about one-third and the latter two-thirds of the weight of the rod
If the moving parts are to be balanced in the plane of the crank,
the following relation applies. Let Wt be the weight of the re-
ciprocating parts, consisting of the piston, pis-
ton rod, crosshead and one-third of the con-
necting rod. Let Wr be the weight of the
rotating parts, consisting of the crank, crank
pin and two-thirds of the connecting rod. Let
W be the weight of the counterbalance at radius
TI and let r be the length of the crank. Then
from Arts. 113 and 126,
Wr, = W rr + f Wtr.
If there are two crank webs as shown in
Fig. 287, half of W must be in the plane of each.
FIG. 287
Problem 1. A piston weighs 300 pounds, the piston rod weighs 100 pounds,
the crosshead weighs 50 pounds, the connecting rod weighs 300 pounds and the
crank pin weighs 30 pounds. Let n = r — 1 foot, and let the arm of the
counterweight balance the crank arm in each case. Determine the counter-
weight for each crank web, the construction being as shown in Fig. 287. If
the cylinder is horizontal, what is the unbalanced vertical force if the engine is
running at a speed of 180 r.p.m.? Ans. 298.5 Ibs. 4050 Ibs.
128. Balancing of Locomotives. Fig. 288 shows a typical
case in the balancing of locomotives, being that of an outside
cylinder locomotive with six drive wheels. The crank is part
of the middle drive wheel, and by means of the side rod the force
from the connecting rod is transferred to the other two wheels.
The motion is one of combined translation and rotation, and the
effects of the reciprocating and rotating parts upon the frame of
the locomotive are the same as if it were running upon a stationary
testing table. The conditions for balancing are the same as those
196 APPLIED MECHANICS [CHAP.X
for the stationary engine discussed in Art. 127. For wheels 1 and
3, the rotating parts to be balanced consist of the crank pin and
one-fourth of the side rod. For wheel 2 the rotating parts consist
of the crank pin, one-half of the side rod and two-thirds of the
connecting rod. The counterweight necessary to balance each
may be computed as in Art. 113.
FIG. 288
The reciprocating parts consist of the piston, piston rod, cross-
head and one-third of the connecting rod. As in the case of
stationary engines it is customary to balance two-thirds of this
reciprocating weight in locomotives. There are two methods of
providing for the balancing of the reciprocating parts. One is
to place all of the counterweight for the reciprocating parts upon
wheel 2, combined with the counterweight for its rotating parts.
The other is to divide the counterweight for the reciprocating
parts equally among the three wheels. With the first method the
large unbalanced vertical force due to the heavy counterweight
on wheel 2 produces a heavy kinetic effect upon the track, so in
general the second method is to be preferred.
Problem 1. For a locomotive of the type of Fig. 288, calculate the counter-
weights necessary to balance each wheel, given the following dimensions and
weights: — wheel diameter, 72 inches; crank arm, 15 inches; weight of side
rod, 500 pounds; weight of boss and crank pin on wheels 1 and 3, 120 pounds;
weight of boss and crank pin on wheel 2, 150 pounds; weight of connecting
rod, 300 pounds; weight of crosshead, 40 pounds; weight of piston rod, 80
pounds; weight of piston, 200 pounds. Use a radius of 28 inches for the
counterweight in wheels 1 and 3, and 27 inches in wheel 2. Divide the counter-
weight for the reciprocating parts equally among the three wheels.
Ans. Wheels 1 and 3, W = 181 Ibs. Wheel 2, W = 387 Ibs.
Problem 2. What is the hammer blow on the track under each wheel of the
locomotive of Problem 1 if it is running with a speed of 60 miles per hour?
Ans. 3100 Ibs.
Problem 3. If the locomotive described in Problem 1, after being properly
balanced, has the connecting rod taken off, but side rod left on, what is the
hammer blow on the track when being pulled by another locomotive at a
speed of 40 miles per hour? Ans. 1380 Ibs. on 1 and 3. 4350 Ibs. on 2.
TRANSLATION AND ROTATION
197
GENERAL PROBLEMS.
Problem 1. The connecting rod shown in Fig. 289 is 5 feet long and the
crank is 1 foot long. Begin at dead center with the connecting rod in the
position AM and locate the position of the instantaneous center of the rod for
each 30° of a half revolution.
FIG. 289
Ans. AM, at A. BN, 3.37 ft. above B. CP, 9.39 ft. above C. DQ, infin-
ity. ER, 7.66 ft. below E. FS, 2.37 ft. below F. GT, at G.
Problem 2. If the engine of Problem 1 is running at 100 r.p.m., find the
velocity of the piston at each 30° point by means of the instantaneous center.
Ans. At A, v = 0. B, 6.15 ft. per sec. C, 10 ft. per sec. D, 10.47 ft. per
sec. E, 8.15 ft. per sec. F, 4.32 ft. per sec. G, 0.
Problem 3. Check the results of Problem 2 by the graphic method.
Problem 4. The connecting rod described in Problem 1 weighs 300 pounds,
its center of gravity is 3 feet from the crosshead end and its moment of inertia
with respect to the axis of the crosshead pin is 107.8. What are the crank pin
and guide reactions for the two positions, as follows: — (1) dead center at head
end, horizontal pressure of crosshead pin 12,000 pounds; (2) at 60° from dead
center, horizontal pressure of crosshead pin 8000 pounds?
Ans. (1) N = 10,895 Ibs.; T = 180 Ibs. upward; NA = 120 Ibs.
(2) N = 7660 Ibs.; T = 273 Ibs. downward; NA = 1377 Ibs.
Problem 5. A steel cylinder 8 inches in diameter and 3 feet long rests with
each end on a horizontal rail normal to the direction of its axis, as shown in
Fig. 290. If static / = 0.25 and kinetic / = 0.20, determine the motion if a
force P = 150 pounds is applied vertically downward to a rope wrapped
around the cylinder at its middle.
Ans. Cylinder rolls to right. Static F = 100 Ibs. a = 6.28 ft. per sec.
per sec.
FIG. 290
Problem 6. Solve Problem 5 if the rope is wrapped through 180° more and
the force is applied vertically upward.
Ans. Cylinder slides to right and rotates clockwise. Kinetic F = 72.5 Ibs
a = 4.55 ft. per sec. per sec. a = 29.2 rad. per sec. per sec.
198 APPLIED MECHANICS [CHAP.X
Problem 7. Solve Problem 5 if the force pulls horizontally to the left on
the rope at the bottom of the cylinder.
Ans. Cylinder slides to left and rotates clockwise. Kinetic F = 102.5 Ibs.
a = 2.98 ft. per sec. per sec. a = 17.9 rad. per sec. per sec.
Problem 8. A plane 10 feet long is inclined at an angle of 15° with the
horizontal. Static / = 0.15 and kinetic / = 0.12. If a cylinder 4 inches in
diameter starts from rest at the top and rolls down, determine the time re-
quired for it to reach the bottom and its linear and angular velocity at the
bottom. Ans. t = 1.865 sec. v = 10.7 ft. per sec. co = 64.2 rad. per sec.
Problem 9. The plane described in Problem 8 is raised to an angle of
24° with the horizontal. Two similar cylinders are released from rest at the
top. The one slides on its base, the other rolls freely. Find the time required
for each cylinder to reach the bottom.
Ans. One slides in 1.44 sec. The other rolls in 1.51 sec.
Problem 10. A plane 20 feet long is inclined at an angle of 30° with the
horizontal. Static/ = 0.25 and kinetic/ = 0.20. Two cylinders are released
from rest at the top and allowed to roll down. One is solid, 6 inches in diam-
eter; the other is hollow, 6 inches outside diameter and 5 inches inside diam-
eter. Find the time for each to reach the bottom.
Ans. 1.93 sec. for solid cylinder. 2.14 sec. for hollow cylinder.
Problem 11. A cast iron cylinder 6 inches in diameter and 6 inches high
rests on end on a -horizontal plane whose coefficient of kinetic friction is 0.20.
If a force of 20 pounds is applied horizontally to a cord wrapped around the
cylinder, find the linear and angular accelerations.
Ans. a = 8.15 ft. per sec. per sec. a = 82.4 rad. per sec. per sec.
Problem 12. Assume that in the engine described in Problem 1, Art. 127,
the cranks are replaced by disks which need no balancing. Instead of the
balance weights shown in Fig. 287, balancing is to be effected by adding one
counterweight to the rim of a large flywheel 30 inches to the left of the plane
of the connecting rod and another to the rim of a small flywheel 24 inches to
the right of the plane of the connecting rod. If the radius of the counter-
weight in the large flywheel is 5 feet and in the small flywheel is 2 feet, find
the weights necessary. Ans. 53 Ibs. 166 Ibs.
Problem 13. The engine described in Problem 12 has two bearings, the
center of each being 10 inches from the center of the connecting rod. What
is the amount of variation in the vertical reaction of each when the engine is
running at 180 r.p.m.? Ans. 4050 Ibs.
CHAPTER XI.
^•-1
FIG. 291
WORK AND ENERGY.
129. Work. The work done by a force is the product of the
force and the distance through which the body upon which it acts
moves in the direction of the force.
In Fig. 291, four forces, FI, W, N and
F are shown acting upon a body resting
upon a horizontal plane surface. Let
the body move through a distance s to
the right as shown. Then forces W and N do no work upon
the body, since the body does not move in the direction of either
force. The work done by force FI is equal to FI X s cos 6, since
s cos 6 is the distance the body moves in the direction of the
force Fj.
Since FI X s cos 9 = FI cos 6 X s, and FI cos 0 is the component of
FI in the direction of s, it may also be stated that the work done by
a force is equal to the product of the distance moved through
by the body and the component of the force in that direction.
If the displacement is in the same direction as the working com-
ponent of the force, the work is positive, as in the case of FI above,
and the work is said to be done by the force. If the displacement
is in the direction opposite to the working component of the force,
the work is negative, as in the case of the frictional force F above.
The work is said to be done against the force, and is equal to — Fs.
If the force is variable, the work done in a small distance ds is
F cos 6 ds as before, 0 being the angle between the direction of the
force and the direction of the motion. The total amount of the
work done by the force is I F cos 6 ds. If the relation befr\
F, 9 and s is known, this expression can be integrated.
For a rotating body, ds = r d<f>, (j> being the angle between the
radius and a fixed axis through the center of rotation, hence
Total Work = f F cos 0 ds = f F cos 0 r d(f>.
199
bween
200 APPLIED MECHANICS [CHAP.XI
But Fr cos 0 is the torque M of the force F about the axis of rota-
tion, 6 being the angle between F and the tangent, so
Total Work = \M d<f>.
If there are several forces producing rotation, M is the resultant
torque. If the torque M is constant, this becomes
Total Work = M<j>,
$ being the total angle in radians described by the radius. The
work done in one revolution is 2 irM = 2 irrF cos 6 = 2 TrrFt, Ft
being the tangential component of F. The normal component Fn
does no work.
The work done on a body against gravity in raising it through
any distance is equal to the product of the weight of the body and
the vertical distance h through which it is raised, irrespective of its
lateral motion. In Fig. 292, the body B is moved along the path
ABC in a vertical plane. The work done against gravity in the
distance ds is equal to Wds cos 0, and the total work is
W f'ds cos B = W Cdy = Wh.
When a body is lowered, gravity does positive work upon i
and the amount of the work is equal to the product of the weight
of the body and its vertical displacement.
,
£ivD JK
V V
FIG. 292 FIG. 293
If a body of weight W is composed of any number of small parts
with weights Wi, w2, ws, etc., which are raised through different
heights, the total amount of work done_is equal to the product of
the entire weight W and the distance H through which the center
of gravity is raised. In Fig. 293, let hi, h2, h3, etc., be the heights
of Wi, wz, ws, etc., above a chosen base line before they are lifted,
and hi, h2f, hzf, etc., their heights above the same plane after being
lifted. Then the total work is given by
ART. 130] WORK AND ENERGY 201
w1 (hi' - h^ + w2 (V - h2) + w3 (h3f - ht) + etc.
= (wjii + wzhz + ^3' + etc.) — (wjii + w2/*2 + wjiz + etc.)
= Wh' - Wh
= WH.
The unit of work is the work of one unit of force through one
unit of distance. In the English system the foot-pound is the
unit most generally used. In the c.g.s. system the unit is the erg,
which is the work done by a force of 1 dyne acting through a dis-
tance of 1 centimeter.
Problem 1. In Fig. 291, W = 1200 pounds and/ = 0.30. If 9 = 30° and
s = 1 inch, find the amount of each force and the work done by it if the motion
is uniform. Ans. Fi = 502 Ibs.; 36.3 ft.-lbs. F = 435 Ibs.; — 36.3 ft.-lbs.
Problem 2. What is the work done against gravity in pulling a 100,000-
Ib. car at uniform speed up a 2 per cent grade a distance of one mile? If train
resistance is 6 pounds per ton, what is the total work done by the drawbar
pull? Ans. 10,560,000 ft.-lbs. 12,144,000 ft.-lbs.
Problem 3. A vertical mine shaft 6 feet square is driven through 120 feet
of clay, 80 feet of shale and 20 feet of sandstone. The material is raised 20
feet above the mouth of the shaft. If clay weighs 100 pounds per cubic foot,
shale 120 pounds per cubic foot and sandstone 150 pounds per cubic foot, what
is the total work done? Ans. 121,608,000 ft.-lbs.
Problem 4. A cable 200 feet long weighing 2 pounds per foot passes over
a pulley with 50 feet hanging on one side and 150 feet on the other. What
work is done against gravity as the pulley is rotated until the middle of the
cable is at the pulley? Ans. 5000 ft.-lbs.
130. Graphical Representation of Work. Since work is the
product of force and displacement, both of which are vector
quantities, the graphical representation of c n
work is made by means of an area. In
Fig. 294, let AB represent the displace-
ment s to some scale, and let AC represent
Work
the magnitude of the force to some scale. 4 Distance — 5
If the force is constant, the area ABDC pIG 294
represents to scale the work done, since it
is the product of AB and AC. If AB represents 4 feet and AC
represents 3 pounds, the area of each small rectangle represents 1
foot-pound of work and the whole area represents 12 foot-pounds
of work.
If the force varies in magnitude, the ordinates will not be the
same height, but the area will still give the work done. By cal-
culus, the area under the curve which has abscissae s and ordinates
202
APPLIED MECHANICS
[CHAP, xi
F is equal to I F ds. By Art. 129 this is the expression for the
work done if the angle G is zero. The simplest case is that in which
the force varies as the distance, increasing from zero to a maxi-
mum or decreasing from an initial maximum to zero. The dia-
gram for this case is a triangle, Fig. 295, in which the ordinate
AB represents to some scale the maximum value of the force F,
which has increased uniformly from zero. The work done is
represented by the triangular area OB A, which is equal to \ AB X
OA. ' In terms of the maximum force FI and the total distance Si,
Work = \ FlSl.
If the relation between F and s is not known, as in the steam
indicator diagram, Fig. 296, the area may be calculated approxi-
mately either by Simpson's One Third Rule or by dividing the
area into a large number of narrow strips and considering each
a trapezoid. An easy method and the one most commonly used
is to measure the area with a planimeter.
.
Distance
FIG. 295
N
Distance
FIG. 296
Let AB, Fig. 296, be the atmosphere line. Then an ordinate
to the top line, as NK, represents to scale the pressure in the
cylinder when the piston was at that point in its travel moving
to the right, so the area A DEB represents the gross work of the
steam. The ordinate to the lower line, as NH, represents the
pressure on the same side of the piston at that point in the return
stroke. This back pressure does negative work, represented by
area ADCHB, so the net work is that done on the forward stroke,
minus that done on the back stroke, represented by area CDKEH.
Problem 1. Draw the diagrams for the Problems of Art. 129.
Problem 2. A force is applied to a 20,000-pound spring to compress it
2$ inches, then released £ inch. (A 20,000-pound spring is one which will be
compressed 1 inch under a static load of 20,000 pounds.) Draw the diagram
and from it compute the gross work, the negative work and the net work.
Ans. 50,625 in.-lbs. gross. 5469 in.-lbs. negative.
131. Energy. Energy is the capacity to do work. If a weight
has capacity to do work on account of its position above a chosen
ART. 133] WORK AND ENERGY 203
datum plane, its energy is called potential energy. If released
from its support it may be made to do positive work in descending
to its zero position. The steam in a boiler and a spring under
compression are said to have potential energy, due to their stressed
condition.
A moving body, by virtue of its velocity, has capacity to do
work as it is brought to rest. This energy of motion is called
kinetic energy.
In addition to these two forms of energy, sometimes called
mechanical energy, several other forms may be classified. Thermal
energy is the capacity to do work on account of the heat possessed
by the body. Chemical energy is the capacity of substances to do
work by combining chemically, as hydrogen and oxygen, which
combine with an explosive force. Electrical energy is the capacity
of a body to do work due to its electrical condition, as, for instance,
a charged storage battery. These other forms of energy are in
reality but different manifestations of kinetic energy, and all the
forms mentioned are mutually convertible.
132. Relation Between Work and Kinetic Energy. Since
kinetic energy is the capacity ,of a body to do work on account of its
motion or velocity, the amount of its kinetic v
energy is necessarily equal to the amount of I ~ l__>/j>
work done by the positive acting force or -, ' — '
forces in producing that velocity. In Fig. 297 r*~* -
let F be the resultant force which acts upon a
particle of mass m to produce the velocity v in the distance s. Then
the work done by the resultant force is equal to I F ds. Since
i Jo
F = Ma and a ds = v dv,
ns /»« r*v
Work = I F ds = / mads = I mvdv = \ mv2.
Jo Jo Jo
So in terms of the velocity the work done is equal to J rat;2, which
is the kinetic energy of the particle.
133. Kinetic Energy of Translation. Forces Constant. The
kinetic energy of a body at any instant is equal to the sum of the
kinetic energies of the particles of which it is composed. In a
motion of translation, each particle of the body has the same
velocity at any instant, so the total kinetic energy is
K.E. = 2 J mv* = i
M being the mass of the whole body.
204 APPLIED MECHANICS [CHAP, xi
Let F, Fig. 298, be the resultant of all the working forces acting
upon a body of mass M as it moves from A to B through the
distance s. Let VQ be its velocity at A and v its velocity at B.
Then
ft
Work = Fds = ° Mvdv.
K ------ £-- ..... ->T
FIG. 298
-/
t/vo
I
J Mv02 is the kinetic energy of the body at A and \ Mv2 is its
kinetic energy at B, so the following general statement may be
made : —
In any motion of translation the positive work done by the resultant
force is equal to the increase in kinetic energy.
C*
If F is constant, I F ds = Fs,
Jo
so
Fs = } Mv2 - \ Mv02.
It is usually simpler in any given problem to consider separately
the work of the several forces acting upon the body instead of the
work of their resultant. Let FI, F2, etc., be the forces and Fiy
FZ'J etc., their components in the direction of the motion of the
body.
Work = jfjy ds + CF2f ds + etc.
If the forces are constant and act through distances Si, S2, etc.,
respectively, in the direction of the motion of the body,
Work = F1fsl + F2's2 + etc.
If some of the forces are resistances, their work is negative.
In any motion of translation the work done by the positive forces
minus the work done by the negative forces is equal to the increase in
kinetic energy. ^\>
This may be written
Positive Work - Negative Work = Final K.E. - Initial K.E.
If the term \ Mv0z is transferred to the other side of the equa-
tion, it may be written
Initial K.E. + Positive Work - Negative Work = Final K.E.
EXAMPLE.
An 80,000-pound car is hauled up a 2 per cent incline by a constant draw-
bar pull of 1000 pounds. If the train resistance is 6 pounds per ton and the
ART. 134J WORK AND ENERGY 205
initial velocity is 20 feet per second, how far up will it go before its velocity is
reduced to 10 feet per second?
Solution: — Initial K.E. = Mv<? = x 2& = 497,000 ft.-lbs.
Positive work = 1000 s, if s is the distance in feet.
Negative work of train resistance = 240 s.
80 000
Negative work of gravity = — — s = 1600 s.
Final K.E. = ^ Mv2 = ^ ^^ X 102 = 124,200 ft.-lbs.
Then 497,000 + 1000 s - 240 s - 1600 s = 124,200.
840 s = 372,800.
s = 444 ft.
Problem 1. A body weighing 100 pounds rests upon a horizontal plane for
which the kinetic coefficient of friction / = 0.10. If a force of 15 pounds is
applied horizontally, what is the velocity of the body 10 feet from the initial
position? Ans. 5.67 ft. per sec.
Problem 2. A body is released from rest at the top of a 45° plane 10 feet
long for which kinetic / = 0.25. What is the velocity at the bottom?
Ans. 18.47 ft. per sec.
Problem 3. If a body is projected with a velocity of 20 feet per second
along a horizontal plane for which kinetic/ = 0.15, how far will it go?
Ans. 41.4ft.
Problem 4. A piston, piston rod and one-third of connecting rod weigh
360 pounds. If the maximum velocity of the piston is 15 feet per second,
what is its kinetic energy? Is this energy lost? Ans. 1259 ft.-lbs.
Problem 5. If the drawbar pull on the car in the Example above is removed
when the velocity is 10 feet per second, how much farther up the incline will
it go? If then allowed to run back, with what velocity will it reach the bottom?
Ans. 67.6 ft. v = 23.6 ft. per sec.
134. Kinetic Energy of Translation. Forces Variable. If
some of the forces acting upon a body during its motion are varia-
ble, the relation between work and change in kinetic energy
becomes
r
Fds =
If F varies with s, it must first be expressed in terms of s if the law
of its variation is known, and then the expression integrated.
If a spring is deformed, the resistance it offers is proportional
to the amount of its deformation. Hence, if a spring is the means
of applying a force to a body, the amount of the force will be some
206 APPLIED MECHANICS [CHAP.XI
constant C multiplied by the distance deformed s, or F = Cs. The
work done by the force is
Cs2 F
Csds = — = ^s.
& £t
This is the product of the average force and the distance.
If one of the forces is steam, working expansively, theoretically
the absolute pressure varies inversely as the distance from the end
of the cylinder, so the amount of the force during expansion is
equal to some constant Ci divided by the distance s from the end
C
of the cylinder, or F = — . An example of each of the cases men-
s
tioned will better illustrate the method of solution.
EXAMPLE 1.
A body weighing 150 pounds falls 8 feet from rest and strikes a 2000-pound
spring. What is the deformation of the spring?
Solution: — The body and spring are shown in Fig. 299. The body is at
rest before starting to fall, hence has zero kinetic energy. When the spring
has its maximum compression, the body is again at rest and has
zero kinetic energy. The resistance of the spring is 2000 X
displacement in inches, or 24,000 X displacement in feet. If s is
the displacement in feet, the resistance of the spring is 24,000 s.
Then
150 X 8 + 150 s - f 24,000 s ds = 0.
Jo
1200 + 150 s = 12,000 s2.
s = 0.3224 ft. = 3.87 in.
The velocity at any point, as when the compression is 1 inch,
may be found by equating the resultant work to the kinetic energy.
150 X 8.083 - (** 24,000 s ds = i ^ v*.
«/0 & OA.4
v = 22 ft. per sec.
EXAMPLE 2.
Fig. 300 (a) represents a steam hammer. The ram and piston weigh 500
pounds. The piston is forced downward by a steam pressure of 10,000 pounds
against an air resistance of 1000 pounds. The stroke s2 is 18 inches. Con-
sidering cut-off when Si = 6 inches, find the velocity of the hammer as it strikes
the metal, assuming pressure X volume = constant for steam after cut-off,
and neglecting clearance volume and friction.
ART. 134]
WORK AND ENERGY
207
ABC
Solution: — The work of the positive forces minus the work of the resisting
forces is equal to the increase in kinetic energy. The weight and the steam
pressure are the positive forces, the at-
mospheric pressure is the resisting force
and its total kinetic energy at striking
is the increase in kinetic energy, since
v0 •= 0.
In Fig. 300 (b), ordinate DB represents
the initial steam pressure Pi, ordinate
BA represents the force of gravity and
ordinate AC represents the resisting
force of the air. Then area DCEFG
represents the resultant work done on
the hammer till the point of striking.
The work of gravity = 500 X 1| =
750 ft.-lbs. The work of steam to cut-
off = PlSl = 10,000 X \ = 5000 ft.-lbs.
The work of steam after cut-off = f 2 P ds, P being the variable pressure. Since
Jsi
the pressure varies inversely as the volume, it also varies inversely as the
distance from the end of the cylinder, or
FIG. 300
P_ ft
Pi «'
p =P^1
5000
x»1.5 fa
I
•Jo.5 5
Then work of steam after cut-off = 5000
= 500010ge
= 5000 loge 3
= 5000 X 2.31ogi03
= 5000 X 2.3 X 0.477
= 5490 ft.-lbs.
The negative work of atmospheric pressure = 1000 X 1.5 = 1500 ft.-lbs.
Work of gravity + work of steam — work of air = £ My2.
750 + 5000 + 5490 - 1500 = \ j~ tf
v = 35.4 ft. per sec.
Problem 1. A body weighing 800 pounds falls 2 inches and strikes upon a
3000-pound spring. What is the deformation? Ans. If inches.
Problem 2. A car weighing 80,000 pounds moving with a speed of 2 feet
per second strikes a bumping post. Assuming the drawbar spring to take all
the compression, what must be its strength in order that the compression shall
not exceed 2 inches? Ans. 29,800-lb. spring.
208 APPLIED MECHANICS [CHAP.XI
Problem 3. A 100,000-pound car moving with a speed of 2 miles per hour
strikes a bumping post. If equipped with a 40,000-pound spring, will all the
shock be taken by the spring if the travel of the spring is 2.5 inches?
Am. No, 2.83 in. required.
Problem 4. A steam hammer has the following dimensions: Weight of
piston and ram, 2000 pounds; diameter of piston, 13 inches; diameter of
piston rod, 2 inches; stroke, 36 inches; absolute steam pressure, 100 pounds
per square inch; air pressure below piston, 15 pounds per square inch; cut-off
at quarter stroke, expansion adiabatic. Neglecting clearance volume and
friction, determine the velocity of striking. Ans. 27 A ft. per sec.
135. Kinetic Energy of Rotation. Let Fig. 301 represent any
body rotating about axis 0 with angular velocity co. The velocity
of any particle of mass dM at a distance p from the axis is pco, and
its kinetic energy is J dMpW. The total kinetic energy of the
body is then / \ dMpW. Since co2 for all the particles at any
instant is the same, and since / dMp2 = /<>,
K.E. of rotation = i/0co2.
The value of 70 may be computed by the regular methods of
integration if the form of the rotating body is regular. If not,
it may be determined by experiment, as in Art. 79.
FIG. 301 FIG. 302
In order to determine the relation between work and kinetic
energy of rotation, the same principle is made use of as in the
case of kinetic energy of translation, Art. 133. Let F, Fig. 302,
acting at distance r from the fixed axis 0 be one of a system of
forces pr9ducing rotation. Then the total work as shown in Art.
129 is
MdO.
Since M = la (Art. 105) and add = codco (Art. 100),
CM de = Cla d0 = JIa do>.
ART. 135] WORK AND ENERGY 209
By integration,
This principle may be stated as follows : —
In any motion of rotation, the positive work is equal to the increase
in kinetic energy.
EXAMPLE 1.
A flywheel and shaft weighing 1800 pounds are rotating in bearings at
80 r.p.m. The shaft is 3 inches in diameter and the radius of gyration of the
shaft and wheel is k = 2.5 feet. If the coefficient of friction / = 0.01, how
long will the wheel rotate?
Solution: — The only force is friction, a resisting force. In one revolution
its work is fN X 2irr, so in n revolutions its work is 0.01 X 1800 X 2x X
0.125 n, or 14.16 n ft.-lbs. The decrease in kinetic energy is the amount of the
initial kinetic energy, or \ 7«2.
/ = MA;2 = ^ X 6.25 = 349.
OZuS
80 X27T
w= ~60~~
o>2 = 70.3.
\ 7w2 = 12,280 ft.-lbs.
Then 14. 16 n = 12,280.
n = 867 revolutions.
Since its acceleration is constant, the average r.p.m. = 40. The time re-
quired for 867 revolutions is given by 867 -j- 40 =21 min. 40 sec.
Problem 1. What is the kinetic energy of a steel cylinder 4 inches in
diameter and 6 feet long when rotating at 240 r.p.m? Ans. 34.95 ft.-lbs.
Problem 2. Compute the kinetic energy of the flywheel shown in Fig. 207
when rotating at 120 r.p.m. Ans. 5896 ft.-lbs.
Problem 3. A cast iron cylinder 1 foot long and 1 foot in diameter is
fastened to a horizontal shaft through its geometric axis. The shaft is 1 inch
in diameter and rests in bearings for which the coefficient of friction/ = 0.015.
If a cord is wrapped around the cylinder and a force of 10 pounds is applied at
the end of the cord during one revolution, what will be
the speed of rotation of the cylinder? (Neglect the
pull on the cord in computing the work of friction.)
Ans. 63.1 r.p.m.
Problem 4. Solve Problem 3 if the 10-pound force is
replaced by a 10-pound weight. Ans. 61.5 r.p.m.
Problem 5. Fig. 303 represents a pulley 2 feet in
diameter fastened to another 4 feet in diameter and
free to rotate about their common geometric axis, 0.
Their combined weight is 200 pounds and their radius
of gyration is 1.5 feet. If a weight of 300 pounds is
hung from a cord wrapped around the smaller cylinder and another of 100
pounds to a cord wrapped around the larger cylinder in the opposite direction,
210 APPLIED MECHANICS [CHAP, xi
find the speed of rotation after the 300-pound weight has moved 2 feet from
rest. Neglect axle friction. Ans. 31.96 r.p.m.
Problem 6. A slender rod 5 feet long weighing 20 pounds is free to rotate
about an axis normal to the rod one foot from the end. If the rod is released
from rest in the vertical position with the center of gravity above the support,
with what velocity will it pass through the horizontal position? What veloc-
ity will it have when it reaches its lowest position?
Ans. 4.72 rad. per sec. 6.68 rad. per sec.
136. Kinetic Energy of Rotation and Translation. A body
which has a motion of combined rotation and translation may be
considered at any instant to be rotating about its instantaneous
axis. As in Art. 135, its kinetic energy at that instant is given
by J Jco2, 7 being the moment of inertia with respect to the instan-
taneous axis. Since the instantaneous axis is not fixed in the
body, it is in general difficult to determine / for the different
instantaneous axes, so a transformation is made.
Since 7 = Ig + Mr2 and v = rco, Ig being the moment of inertia
of the body with respect to the gravity axis parallel to the in-
stantaneous axis, r the distance between them and v the absolute
velocity of the center of gravity,
K.E. = i 7co2 = \ 10? + \
The term \ Mv* is the expression for the kinetic energy the body
would have if moving in translation with velocity v, and the term
| 700>2 is the kinetic energy it would have if rotating about a fixed
axis through the center of gravity parallel to the instantaneous
axis. Hence the kinetic energy of a body with any plane motion
is equal to the sum of the kinetic energies of translation and of
rotation about the center of gravity.
EXAMPLE.
A wheel 2 feet in diameter weighing 100 pounds starts from rest and rolls
freely down a 30° incline. If the plane is 10 feet long and the radius of gyra-
tion k = 0.8 foot, what is the velocity of the center of the wheel as it reaches
the bottom? What is the kinetic energy of rotation? Of translation ?
Solution: — The vertical height h = 5 feet, so the work of gravity is 100 X
5 ft.-lbs.
The frictional force F and the normal resistance N do no work, since their
point of application is at rest.
7 = MA:2, so
500 = \ Mi* + \ MkW.
ART. 137] WORK AND ENERGY 211
100
For free rolling, v = ru = o>; also M = ^~^'
v = 14.02 ft. per sec.
K.E. of rotation = \ Af Afto2 = 195 ft.-lbs.
K.E. of translation = \ Mv2 = 305 ft.-lbs.
Problem 1. Solve for the velocity of the disk in Problem 1, Art. 121, by
the method of this article.
Problem 2. A solid sphere 2 feet in diameter weighing 100 pounds rolls
freely down a 30° incline 10 feet long. Find its linear velocity at the bottom.
Ans. 15.16 ft. per sec.
Problem 3. What is the kinetic energy of a cast iron disk 2 feet in diameter
and 4 inches thick which rolls along a level floor with a velocity of 10 feet per
second? Ans. 1098 ft.-lbs.
Problem 4. The connecting rod of Problem 2, Art. 79, and Problem 1,
Art. 123, weighed 267 pounds and its center of gravity was 48.5 inches from
the crosshead pin. By experiment, 7 with respect to the axis of the crosshead
pin was found to be 157. The crosshead to which it was attached weighed
100 pounds, the piston rod weighed 70 pounds and the piston weighed 290
pounds. Find the total kinetic energy of the piston, piston rod, crosshead and
connecting rod at dead center (head end), at 45° and at 90°.
Ans. 775 ft.-lbs. 2774 ft.-lbs. 4015 ft.-lbs.
Problem 6. A freight car weighing 80,000 pounds has four pairs of wheels
like those described in Problem 1, Art. 79. Find the percentage of error if the
rotational component of the kinetic energy of the wheels is neglected in com-
puting the total kinetic energy of the car. Ans. 0.59 of 1 per cent.
137. Work Lost in Friction. Friction is the great reducing
agent by means of which kinetic energy is dissipated in the form
of heat. When two surfaces move over each other the mutual
friction reduces the total kinetic energy by an amount equal to
the product of the frictional force and the relative motion. Let
the crosshead of a locomotive have an average pressure on its
guides of 500 pounds and let the coefficient of friction / = 0.01.
If the crank is 1 foot long, the relative distance which the cross-
head moves over the guides during the forward stroke is 2 feet.
The work lost in friction between the crosshead and guides is
then 0.01 X 500 X 2 = 10 ft.-lbs. This cannot be recovered in
the form of useful work in another part of the motion as in the
case of the work required for accelerating the piston, for the friction
changes direction and the same amount more is lost during the
return stroke.
The brake shoe testing machine, Fig. 304, consists of a heavy
drum A rigidly fastened to an axle 0, to which is also fastened a
car wheel B. By means of the levers C and D the weight P
212 APPLIED MECHANICS [CHAP.XI
applies a normal pressure to the rim of the wheel through the brake
shoe E. The weights and dimensions are known, so the kinetic
energy may be computed when the angular velocity is known.
The weight of the rotating drum is 12,600 pounds, practically the
same as i of a 100,000-pound car.
or the part supported by one wheel.
The material is so arranged that
its rotary kinetic energy is the
same as the translatory kinetic
energy of the same weight would
be if it had the same speed as the
FIG 3Q4 rim of the wheel. The require-
ment for this is that i /or = £ Mv2.
Since I = Mk2 and v = no, it is necessary that k = r, so the drum
is built in such a way as to make k = 1.375 feet, the radius of
the standard car wheel. The brake shoe on the testing machine
is under the same conditions as in service.
The axle is connected to an engine by means of a clutch so that
any desired speed may be given to it, after which the engine may
be disconnected. The drum and wheel will then rotate freely
until the weight of P is applied which presses the brake shoe E
against the wheel. The arrangement of the levers is such that a
weight P at the end of the lever produces a normal pressure of
24 P at E. In addition, the weight of the levers themselves
produces a normal pressure of 1230 pounds. Let the total
normal force be N. Then the frictional drag of the shoe upon the
wheel is fN, f being the kinetic coefficient of friction, and the
work of the frictional force is fNir yf ft.-lbs. in one revolution.
Some work is done also by the frictional force at the bearings.
The pressure here is 12,600 + AT. The axle is 7 inches in diameter
and if fi is the coefficient of friction the work lost in one revolution
will be /i (12,600 + N) *• A ft.-lbs. The work of friction at the
two points dissipates the kinetic energy of the drum and wheel.
That at the brake shoe is, of course, much the larger. The
kinetic energy is transformed into heat, sometimes making the
surface of the shoe red hot.
If n is the number of revolutions of the wheel, the work-energy
equation for the motion becomes
K.E. = work of friction.
} /co2 = fNwn X f } + /i (12,600 -f- N) TTU X A
ART. 138] WORK AND ENERGY 213
Problem 1. Use / = 0.25 and /i = 0.005 for the brake shoe machine
described above. If the rim speed of the wheel is 60 miles per hour, what
weight P will bring the wheel to rest in 100 revolutions? Ans. 237 Ibs.
Problem 2. If the rim speed of the wheel on the testing machine is 75 miles
per hour and with 300 pounds at P is brought to rest in 120 revolutions, what
is the value of/? Use the same value for/i as above. Ans. / = 0.268.
Problem 3. If force P in Problem 2 is 600 pounds, in what distance will
the wheel be stopped when running with a rim speed of 90 miles per hour?
Ans. 809ft.
138. Braking of Trains. A train which is running at a high
rate of speed has a large amount of kinetic energy which has been
given to it by the work of the steam in the cylinders, or if on a
down grade, by gravity also. When the train is to be stopped, all
KStaticF,
W,
.(a) Wheel Rolling (b)Wheel Skidding
FIG. 305
this kinetic energy must be used up again in work. The usual
method is to press brake shoes against the rims of the wheels and
so transform the kinetic energy of the train into heat at the
rubbing surfaces. The force of friction does work of retardation
equal to fNs, f being the coefficient of friction, N the normal
pressure of the brake shoe on the wheel and s the distance traveled
by the rim of the wheel relative to the brake. The action is a
tendency to check the rotation of the wheel, so that a backward
static frictional force is developed at the point of contact of the
wheel and the rail. This force of the rail on the wheel is the one
which actually stops the train, but it does no work since its point
of application does not move in the direction of the force. The
maximum braking force is exerted when skidding of the wheel is
impending, but the wheel is still rolling, as in Fig. 305 (a), for then
the limiting or maximum value of the static friction at the rail is
induced. If the friction at the axle is neglected, the equation of
moments about the axle shows that the two frictional forces FI
and F are equal when the wheel is under static conditions.
214
APPLIED MECHANICS
[CHAP, xi
If now the normal force on the brake shoe is slightly increased,
the frictional force at the shoe will become slightly greater and the
wheel will skid, as in Fig. 305 (b). There is now kinetic friction
between the wheel and the rail which is less than static friction,
so resistance to motion is less. The work is being done at the
point of contact of the wheel and the rail instead of at the surface
of the brake shoe as before, while the brake shoe does no work and
is not worn. Skidding of the wheels, besides being much less
efficient in stopping the train, causes injurious flat spots to be
worn on the wheel.
The coefficient of friction between brake shoes and car wheels
is extremely variable on account of the following factors : —
material of the shoe, material of the wheel, initial speed of the
train, speed of the wheel at the time considered and weather con-
ditions. Just before the wheel is stopped the coefficient is con-
siderably higher than the average for the stop. At high speeds
the coefficient is much less than at low speeds, due to the heating
of the material at the rubbing surfaces.
The following table gives average results from some M. C. B.
Association tests upon several different kinds of brake shoes, at
two different speeds.
Speed (mi./hr.)
Average/
Final/
40
65
20.5
10.3
32.6
18.0
It is seen that both the average and the final values of / when the
initial speed is 65 miles per hour are much less than they are when
the initial speed is 40 miles per hour. This is due to the fact that
at 65 miles per hour there is more than two and one-half times
as much kinetic energy to be used up at the rubbing surfaces.
The surfaces are heated more and their gripping qualities lessened.
When stopping trains at high speeds it is customary to apply the
brakes at first with a heavy pressure until the speed is partially
reduced, then to release and apply again with a less pressure in
order to avoid skidding as the train comes to rest.
Since the average value of the final kinetic coefficient of friction
for low speeds is as high as the value of the static coefficient of
friction between the wheel and the rail, the normal pressure on
ART. 139] WORK AND ENERGY 215
the brake shoe should not be greater than the weight carried by
the wheel if skidding of the wheel is to be avoided.
Problem 1. A wheel carries a load of 10,000 pounds. If the static coeffi-
cient of friction fi between the wheel and the rail is 0.30 and the maximum
value of the kinetic coefficient of friction / between the wheel and the brake
shoe is 0.40, what is the maximum allowable normal brake shoe pressure?
An*. N = 7500 Ibs.
Problem 2. A 1000-ton train while running at a speed of 30 miles per hour
down a 0.5 per cent grade has brakes applied so that the train is brought to
rest in 1200 feet. What is the total induced resisting force required, train
resistance being 10 pounds per ton? Ans. 50,000 Ibs.
Problem 3. If the coefficient of static friction between wheel and rail is
0.25, what would be the shortest distance in which the train of Problem 2 could
be stopped? Ans. 123 ft.
139. Power. Power is the rate of doing work, or the amount
of work done per unit of time. If a weight of 100 pounds is lifted
10 feet, the work done is the same whether it is lifted in 1 second
or in 5 seconds. The power required, however, is different. In
the first case the power is 1000 foot-pounds per second, while
in the second it is 200 foot-pounds per second.
The unit of power is the unit of work developed in the unit of
time. In the English system it is, therefore, the foot-pound per
second (abbreviated ft.-lb. per sec.) This is too small a unit for
some engineering work, so the larger unit of the horsepower is also
used. The horsepower is 550 foot-pounds of work per second, or
33,000 foot-pounds of work per minute. In electrical work the
unit of power commonly used is the watt, which is 107 ergs per
second, or the kilowatt, which is 1000 watts. 1 h.p. = 0.746 kw.,
or approximately f kw. If a force F moves through distance ds
ds
in dt time, its rate of doing work or its power is F -r = Fv. So if
F is in pounds and v in feet per second,
Fv
Horsepower = -r^--
Due to friction a certain amount of the energy supplied to a
machine is lost, so the amount delivered by it, the output, is less
than that delivered to it, the input. The ratio of the output to
the input for a given length of time is called the mechanical
efficiency.
_,_ . Output
Efficiency = --
216 APPLIED MECHANICS [CHAP, xi
Problem 1. If a hoisting engine lifts a mine cage weighing 600 pounds 500
feet in one minute, what horsepower is expended? If an indicator shows that
10.5 horsepower is being developed, what is the efficiency of the engine and
hoist? Ans. 9.09 h.p. 86.5 per cent.
Problem 2. Find the amount of useful work done by a pump which dis-
charges 300 gallons of water per minute into a tank 200 feet above the intake.
Ans. 15.22 h.p.
Problem 3. The driving side of a belt has 800 pounds tension and the slack
side has 350 pounds. If the pulley is 2 feet in diameter and has a speed of
240 r.p.m., what horsepower is being transmitted? If it is driving a dynamo
which has an efficiency of 85 per cent, how many kilowatts are being delivered?
Ans. 20.6 h.p. 13.08 kw.
Problem 4. An engine hoists 6 cubic feet of concrete and a 150-pound
bucket 15 feet in 10 seconds. If concrete weighs 125 pounds per cubic foot,
what horsepower is the engine delivering? Ans. 2.46 h.p.
140. Water Power. If a stream of water has a cross-sectional
area of A square feet and a velocity of v feet per second, the cubic
feet of water flowing past any point in one second is Av and its
weight is 62.5 Av. Therefore, the kinetic energy per second, or
power, is
1 W
± IL 3,2 = 0.97 Av3 foot-pounds.
« Q
If the velocity of the water as it reaches any certain point could
be entirely destroyed in doing useful work, the horsepower gen-
0 97 Av*
erated would be — ^^ —
If a flowing stream has a vertical fall of h feet and W is the weight
of water flowing per second, the number of foot-pounds of work
Wk
per second is Wh. Then h.p. = r^-
Problem 1. A river has a cross-sectional area of 90 square feet and a
velocity of 8 feet per second above a fall which has a drop of 6 feet. What is
the theoretical horsepower that could be developed? Ans. 572 h.p.
Problem 2. A 2-inch nozzle discharges water under a head of 1200 feet.
The stream impinges upon a Pelton wheel which has an efficiency of 80 per
cent, and the Pelton wheel is connected to generators which have an efficiency
of 90 per cent. How many kilowatts can be delivered at the switchboard?
Ans. 444 kw.
Problem 3. If a volume of 300 cubic feet of water per second under a head
of 40 feet flows through turbines which have an efficiency of 85 per cent, what
horsepower will they deliver? Ans. 1160 h.p.
141. Steam Engine Indicator. The steam engine indicator is
an instrument for recording graphically the steam pressure and
ART. 142] WORK AND ENERGY 217
piston travel of a steam engine. In Fig. 306, OX is the axis of
zero absolute pressure. AB is the line showing atmospheric pres-
sure, and its length represents to some scale the piston travel.
The ordinate FA represents to some scale
the steam pressure when the piston was at
the corresponding point in its stroke. As
shown in Art. 130, the area GCDE repre-
sents the net work done by the steam. v
This area divided by the length AB will
give the average ordinate, which multiplied
by the scale of the spring will give the mean effective pressure.
This is the pressure which, if exerted through the whole stroke,
would have done the same amount of work.
. Let P represent the mean effective pressure in pounds per
square inch, I the length of the cylinder in feet, a the area of the
piston in square inches and n the number of revolutions per
minute made by the flywheel. Then Pa is the total pressure in
pounds and Pla is the work done in one revolution by the steam
on one side of the piston. Plan is the work done in one minute
and „ is the horsepower generated. This is called the indi-
cated horsepower (I.H.P.). If the piston rod extends both ways
from the piston so that the areas are the same and the mean
effective pressures are the same, the total power generated in
the cylinder is ^-^^ - If, as is commonly the case, the pressures
and areas are different, the horsepowers of the two ends are com-
puted separately and added.
Problem 1. The mean effective pressure of an engine is 80 pounds per
square inch; the crank is 12 inches long; the cylinder is 7 inches in diameter;
the piston rod is 1.5 inches in diameter and extends entirely through the
cylinder. If the engine is running at 150 r.p.m., what is the indicated horse-
power? Ans. 53.4 h.p.
Problem 2. At the crank end of the cylinder of an engine the mean effective
pressure is 115 pounds per square inch and at the head end is 110 pounds per
square inch. The length of the crank is 1 foot, the diameter of the cylinder
is 1 foot and the diameter of the piston rod is 2 inches. The piston rod
extends only one way from the piston. If the speed of the engine is 90 r.p.m.,
what horsepower is it developing? Ans. 137 h.p.
142. Absorption Dynamometer. An absorption dynamometer
is an instrument for measuring the output of power of prime
218
APPLIED MECHANICS
[CHAP, xi
FIG. 307
movers, such as steam engines, electric motors and water wheels.
It absorbs all the energy generated and transforms it into heat of
friction. The most common form of absorption dynamometer
is the Prony brake, Figs. 307 and 308.
The simple construction of Fig. 307 is
best for small, high-speed machines, as
motors and small gas engines. By
tightening the hand wheel J9, the
blocks of which the brake is composed
are clamped against the wheel and the
friction developed tends to turn the
brake around with the wheel. This
tendency is resisted by the pull of the
weight W, hinged at C. The hand wheel B is tightened until all
of the work done by the prime mover is used up in the heat of
friction at the rim.
Let F be the total friction generated. Then the work absorbed
in one revolution is F X 2 irri, and if n is the number of revolutions
per minute the horsepower is given by
_ F X 2 Trrin
33,000
But Fri = Tr2, by moments about 0, and TV3 = TFr4 sin 6 by
moments about C, so
_ 2 irnWrzn sin 0
33,000 n
Instead of being graduated in degrees, the arc may be graduated
to read T directly, or, more simply, the frictional force F.
The brake shown in Fig. 308 is used for larger sizes of flywheels.
FIG. 308
It is composed of small blocks of wood fastened to a strap encir-
cling the wheel and attached to a V-shaped lever arm, CD A. To
ART. 143]
WORK AND ENERGY
219
apply the brake, the ends of the band at E are drawn together
by turning the hand wheel B. As before,
Fn = Pr2,
2 irnPr2
so
h.p.
33,000
The force P is the net force due to friction alone after the weight
of the brake arm has been balanced. It is usually measured by
resting the lever arm on a platform scale, or by suspending it from
a spring scale above.
Since some of the work done by the steam in the cylinder is
used up in friction of the moving parts of the engine, the brake
horsepower will necessarily be less than the indicated horsepower.
The mechanical efficiency is the ratio of the brake horsepower to
the indicated horsepower, or
B.H.P.
Efficiency = IHp '
Problem 1. In a brake of the style of Fig. 307, r\ = 3 inches, r2 = 2 feet,
r3 = 3 inches, r4 = 14 inches and W = 10 pounds. If balanced so that the
pointer is vertical with no load, at what angles should the calibration marks
be placed for F = 50 pounds; F = 100 pounds; F = 150 pounds; F = 200
pounds? Ans. 7° 42'; 15° 33'; 23° 42'; 32° 25'.
Problem 2. If a motor which is being tested by the brake described in
Problem 1 is running at 500 r.p.m. and the pointer reads 170 pounds, what is
the brake horsepower? Ans. 4.05 h.p.
Problem 3. A brake of the style of Fig. 308 has the radius r\ = 2 feet and
r2 = 6 feet. If when a certain engine is being tested P = 410 pounds and
n = 180, what horsepower is being developed? Ans. 84.3 h.p.
Problem 4. If the engine referred to in Problem 3 has a cylinder 16 inches
in diameter, piston rod 3 inches in diameter (extending only one way from the
piston), 24-inch stroke and mean effective pressure of 21 pounds per square
inch, what is the mechanical efficiency of the engine? Ans, 93.2 per cent.
WV '
FIG. 309
143. Band Brakes. Fig. 309 shows a simple form of band
brake, such as is used on hoisting engines. The band is attached
220 APPLIED MECHANICS [CHAP, xi
at C and passes around the wheel to the lever at B. The lever
is hinged at A and if force is applied downward at the end, the
band is tightened and the friction retards the motion of the
wheel inside the band. If the weight W is to be lowered at a
uniform rate of speed, the work done by gravity upon the weight
must equal the work done by the frictional force T2 — TI on the
rim of the brake wheel, or
2<irr2W = (T2- Ti)2vri.
From Art. 64,
T2 = 7V»
f being the kinetic coefficient of friction and 0 the angle of contact.
By moments about point A,
PI = Tia.
From these three equations the force P required to lower the
weight W may be determined.
The band brake with modifications is used on automobiles. If
an automobile of weight W is moving with a speed of v, its kinetic
1 W
energy is -= — v2. In bringing the automobile to rest in a short
distance by means of the brakes, this kinetic energy is used up
chiefly in the work of friction on the rubbing surfaces. If n is the
number of revolutions the wheel makes until it is brought to rest,
the relation between work and kinetic energy may be written
z Q
T2 and TI are the sums of the tensions on the tight and loose ends
respectively of the two brakes.
Problem 1. In Fig. 309, n = 18 inches, r2 = 24 inches, W = 1200 pounds,
a = 5 inches, I = 48 inches and / = 0.15. What force P is required in order
that the weight may descend uniformly? Ans. 162 Ibs.
Problem 2. If P = 150 pounds on the band brake of Problem 1, with what
velocity will the weight W pass a point 50 feet below the starting point?
Ans. 15.54 ft. per sec.
Problem 3. If in Fig. 309 the rope sustaining W is wrapped the other way
on the drum so that W descends on the left side, what force P will be required
to allow it to descend uniformly? Ans. 329 Ibs.
Problem 4. The weight of an automobile is 3000 pounds. The diameter
of its wheels is 36 inches and that of its brake rims is 16 inches. If equipped
with brakes similar to that of Fig. 309 for which a = 1 inch, I = 6 inches and
/ = 0.40, in what distance can it be brought to rest from a speed of 30 miles
per hour by a total pressure P = 40 pounds if the wheels do not skid?
Ans. 151 ft.
WORK AND ENERGY 221
GENERAL PROBLEMS.
Problem 1. Find the amount of work done in elevating the clay from a pit
30 feet in diameter and 50 feet deep if the clay weighs 100 pounds per cubic
foot and is lifted 6 feet above the top of the pit. Ans. 109,563,000 ft.-lbs.
Problem 2. What is the work done against gravity in filling a standpipe
80 feet high and 16 feet in diameter if the water is drawn from deep wells whose
average level is 120 feet below the base of the standpipe?
Ans. 160,848,000 ft.-lbs.
Problem 3. If the standpipe of Problem 2 is to be filled in 6 hours, no water
being drawn out meanwhile, what horsepower must the engine develop?
Ans. 13.56 h.p.
Problem 4. A fire engine takes water from the surface of a lake 20 feet
below its own level and delivers it from a nozzle 2 inches in diameter with a
velocity of 100 feet per second. What horsepower is required?
Ans. 43.5 h.p.
'Problem 6. A tank 8 feet long, 6 feet wide and 4 feet deep is half full of
water. How many foot-pounds of work are required to raise all of the water
over the top of the tank? Ans. 18,000 ft.-lbs.
Problem 6. A block of granite 8 feet long, 3 feet wide and 3 feet high is
lying on its side. How many foot-pounds of work are required to tip it up
on end? (Granite weighs 160 Ibs. per cu. ft.) Ans. 31,960 ft.-lbs.
Problem 7. How much work is done in winding up a cable 500 feet long
which weighs 2 pounds per foot? Ans. 250,000 ft.-lbs.
Problem 8. A mine cage weighs 500 pounds, empty coal car weighs 200
pounds and the cable weighs 1 pound per foot. What work is done in raising
500 pounds of coal from the 200-foot level to the 50-foot level?
Ans. 198,750 ft.-lbs.
Problem 9. A 400-ton train is running at a speed of 60 miles per hour on a
level track. If train resistance is 17 pounds per ton, what is the drawbar pull?
What horsepower is the engine developing? Ans. 6800 Ibs. 1088 h.p.
Problem 10. A 1000-ton train attains a speed of 30 miles per hour in 1 mile
on a level track with a constant drawbar pull. If train resistance is considered
constant and equal to 8 pounds per ton, what is the drawbar pull? What is
the horsepower developed when the speed is 30 miles per hour?
Ans. 19,390 Ibs. 1550 h.p.
Problem 11. If the locomotive of Problem 10 is pulling the same train up
a 0.25 per cent grade and is exerting the'same drawbar pull, in what dis-
tance will it attain a speed of 30 miles per hour?
Ans. 9400ft.
Problem 12. An ore car weighing 500 pounds
is pulled up a 30° incline by means of a counter-
weight as shown in Fig. 310. If the counter-
weight weighs 600 pounds and the car resistance
is 15 pounds, what velocity will the car have
after moving 100 feet from rest?
Ans. 18.6 ft. per sec. FIG. 310
222 APPLIED MECHANICS [CHAP, xi
Problem 13. A coal incline is 500 feet long and 30 feet high. A car
weighing 60,000 pounds runs down the incline and out on level track. If car
resistance is 6 pounds per ton, how far out on the level will it run?
Ans. 9500 ft.
Problem 14. If in a steam hammer of the same dimensions as given in
Problem 4, Art. 134, the steam is cut off and released at the top end of the
cylinder when the piston is at quarter stroke and at the same time is admitted
at boiler pressure below the piston, how far down will the hammer move?
Ans. 1.83 ft.
Problem 15. In the hammer described in Problem 14, at what point must
cut-off and release above, and admission below be made in order that the ham-
mer just touches the anvil? Ans. 1.23ft.
Problem 16. If in the hammer described in Problem 14 the cut-off is made
at quarter stroke but release above and admission below is made at half stroke,
with what velocity will the hammer strike the anvil? Ans. 9.81 ft. per sec.
Problem 17. A Pelton wheel is driven by a jet of water 1 inch in diameter
under a head of 280 feet. If the efficiency of the wheel is 80 per cent, what
horsepower will be generated? If the wheel is directly connected to a genera-
tor which has 90 per cent efficiency, how many kilowatts will be delivered?
Ans. 18.64 h.p. 12.52 kw.
Problem 18. The steam indicator card for the head end of the cylinder
of a steam engine had an area of 3.27 square inches and for the crank end 3.21
square inches. Length of atmosphere line was 3.25 inches; scale of the spring,
40 pounds per inch; length of stroke, 24 inches; diameter of piston, 10 inches;
diameter of piston rod, 1.25 inches. If the engine is running at 120 r.p.m.,
what is the indicated horsepower? Ans. 45.2 h.p.
Problem 19. A hoisting engine is lifting one ton of ore per minute from a
steamer's hold 40 feet deep. If the efficiency of the hoisting apparatus is
75 per cent and that of the engine is 85 per cent, what is the indicated horse-
power? Ans. 3.8 h.p.
Problem 20. What horsepower is being developed if a 1000-ton train is
being hauled up a 0.5 per cent grade at a speed of 30 miles per hour, the train
resistance being 10 pounds per ton? Ans. 1600 h.p.
Problem 21. If the train described in Problem 20 has the steam shut off
and the brakes applied with their maximum effect (/ = 0.25), how far will the
train run? Ans. 116 ft.
CHAPTER XII.
IMPULSE, MOMENTUM AND IMPACT.
144. Impulse and Momentum. The effect of a force may be
given in terms of the product of force and distance, which is called
work, or the product of force and time, which is called the impulse
of the force. If a force F is constant both in magnitude and direc-
tion during time t, the impulse is Ft. If F varies in magnitude,
the impulse for the infinitesimal time dt is F dt, and the impulse
for any time t is given by I F dt. If the relation of F and t is
i/O
known, the integration may be performed. Sometimes, even
though the relation between F and t is not known, the quantity
I F dt can be eliminated between two simultaneous equations
Jo
containing it.
Impulse, like force, is a vector quantity, and has the same direc-
tion and position as the force factor.
The resultant impulse of a force which varies in direction is the
vector sum of the separate component impulses. Consider a
resultant force F applied to a body for t seconds, then suddenly
reversed and applied for a succeeding t seconds. It is evident
that the vectorial sum of the impulses is zero, since they are of
the same numerical value and of opposite sign.
If in the preceding case the force should be changed only through
90°, the resultant impulse would be Ft V2 in amount, at an angle
of 45° with the direction of either component impulse. In any
case in which the force varies in direction, / F dt must be vectorial.
The momentum of a body is the product of its mass and velocity,
Mv. It is sometimes called the quantity of motion. Momentum,
like velocity, is a vector quantity having definite direction and
position. Like other vector quantities, both impulse and momen-
tum may be resolved into components or combined into resultants.
The unit of impulse is the impulse of a unit force acting for a
unit of time. In the English system this is the pound-second.
223
224 APPLIED MECHANICS [CHAP, xn
The unit of momentum is the momentum of a unit mass moving
with unit velocity. In the English system the dimensions of this
W
unit are obtained as follows. Since M — — , W being in pounds
j . feet ,, . . . . pounds X seconds2
and g in - — pr, M is in units of - — . Velocity
seconds2 feet
., f feet , _ . . pounds X seconds2
v is in units of -p, so Mv is in units of — - X
seconds' feet
feet
pounds X seconds. The dimensions of the unit of momen-
seconds
turn are, therefore, the same as those of the unit of impulse.
145. Relation Between Impulse and Momentum. Let F be
the resultant force acting upon a body of mass M to produce an
acceleration a. Then F = Ma. If F is constant, a is constant
and equals V°. The equation then becomes
Ft'=Mv-Mv0.
If F varies in magnitude, F = M-rr, since a = -7- , and
ut Ctt
Fdt = Mdv.
For time t, if VQ is the velocity when t = 0 and v the velocity after
time t,
f'fdl = f*
I/O V%
/
%J Q
f = \ Mdv.
Vdt = Mv- Mvc
The general statement of this relation may be made as follows:
During any period of time the impulse of the resultant force acting
upon a body is equal to its change in momentum.
By means of the above relation, problems involving force, mass,
velocity and time may be solved directly instead of with the double
set of equations between force, mass and acceleration, and velocity,
acceleration and time.
EXAMPLE 1.
A body falls freely for 3 seconds. What is its velocity if it started from
rest?
Solution: — The force is the weight W, so the impulse is 3 W. The change
in momentum is the same as the final momentum, or Mv.
AKT. 146] IMPULSE, MOMENTUM AND IMPACT 225
Then
v = 96.6 ft. per sec.
EXAMPLE 2.
A 500-pound body initially at rest is acted upon for 10 seconds by a variable
working force F which is equal to 100 VI, and also by a variable resisting
frictional force FI which is approximately equal to 20 — t during that time.
What is its velocity at the end of 10 seconds?
Solution: — Impulse = change in momentum.
/«i
jr
500
v.
v = 126ft. per sec.
Problem 1. A body is projected downward with a velocity of 80 feet per
second. What is its velocity 2 seconds later? Ans. 144.4 ft. per sec.
Problem 2. If a 100-pound weight on a level floor has a horizontal force of
15 pounds acting upon it for 10 seconds and a frictional force which is equal to
10 — fi during that time, what will be its velocity if it starts from rest?
Ans. 21.3 ft. per sec.
146. Conservation of Linear Momentum. In any mutual
action between two bodies or two parts of the same body, the
mutual forces are always equal and opposite, by Newton's Third
Law of Motion. Since the time of contact is necessarily the same,
the impulses of the mutual forces are equal in value and opposite in
direction, hence neutralize each other. Then if there is no resultant
force acting which is external to the two bodies, the sum of the
momenta before the action is equal to the sum of the momenta
after the action, since for the whole system I F dt = 0. If MI and
M2 are the masses of two bodies, vi and v2 their velocities before
contact and t;/ and vj their velocities after contact, then
+ Mzv2 = MiVi + M2v2'.
Though there is no loss of momentum in any mutual action
between two bodies, there is always a loss in kinetic energy due
to the heat generated at the point of contact.
The direction of vi should always be considered positive. If v2
is in the opposite direction, it must be used as negative.
226 APPLIED MECHANICS [CHAP.XII
EXAMPLE.
A 50-pound shot is fired from a gun which weighs 20,000 pounds. If its
muzzle velocity is 1200 feet per second, what is the initial I ackward velocity
of the gun? If the recoil is against a constant force of 3000 pounds, how soon
will the gun be brought to rest? What distance does it recoil? What is the
kinetic energy of each?
Solution: — The momentum of the shot and the gun before the shot is
fired is equal to zero, so the sum of the momenta after the shot is hred must
also equal zero. Therefore,
Momentum of shot forward — momentum of gun backward = 0.
50 s 1onn 20,000
v = 3 ft. per sec.
The impulse of the resisting force is equal to the momentum of the gun, so
t = 0.621 sec.
Since the resisting force is constant, the motion is one with uniform ac-
celeration and the distance is equal to the product of the average velocity and
the time. The average velocity is £ X 3 = 1.5, so
s = 1.5 X 0.621 = 0.932 ft.
The kinetic energy 01 the shot is
1 "SO
2 32~2 X 12°°2 = 1'120>000 ft--lbs-
The kinetic energy of the gun is
\ ?Mf x 32 = 279° ft-ibs-
It will be noticed that though the momentum of the shot is numerically equal
to that of the gun, its kinetic energy is 400 times as great.
Problem 1. A man weighing 150 pounds jumps with a velocity of 10 feet
per second into a boat which weighs 120 pounds and which is at rest. What
will be the initial velocity of the boat? What will be the loss in kinetic energy?
Ans. 5.55 ft. per sec. 104 ft.-lbs.
Problem 2. A gun weighing 160,000 pounds fires a projectile weighing
1000 pounds with a velocity of 1500 feet per second. With what initial
velocity will the gun recoil? How far will it recoil if resisted by a nest of
springs which would be compressed 1 inch by a force of 200,000 pounds?
What is the kinetic energy of the projectile and of the gun respectively?
Ans. 9.37 ft. per sec. Scinches. 35,000,000 ft.-lbs. 218,000 ft.-lbs.
147. Angular Impulse and Angular Momentum. In Fig. 311,
let F be one of a system of forces acting upon the body shown, and
let d be the perpendicular distance from 0 to the line of action
of F. The impulse of the force F during time dt is F dt and this
ART. 147] IMPULSE, MOMENTUM AND IMPACT 227
impulse is a vector quantity having direction and line of action as
shown. The moment of this impulse about the axis 0 is d X Fdt
and the summation of these moments of impulse, I d X F dt,
is called the moment of impulse or angular impulse of the force
upon the body.
FIG. 311 FIG. 312
In Fig. 312, let dM be the mass of any particle at distance p
from the axis of revolution 0. If the angular velocity of the body
is o>, the tangential velocity of the mass dM is po> and its momen-
tum is dMpu. This momentum is a vector quantity whose posi-
tion is through dM in the direction of the velocity of dM, normal
to p. The moment of the elementary momentum about the axis 0
is dMp2o) and is called the moment of momentum of the particle.
The summation of all these elementary moments of momentum is
called the moment of momentum or angular momentum of the body.
J
dMp2co = co dM = I0(o.
From Art. 105,
^
at
xFdt = Idu.
fd
*J
fd X Fdt = f M dt = Ico -
The statement of this relation is as follows :
In any motion of rotation, the sum of the angular impulses of the
forces acting is equal to the increase in angular momentum.
Both angular impulse and angular momentum are vector quan-
tities and are represented graphically by vectors parallel to their
axes of rotation. The convention for sign is the same as that
used in connection with couples, as explained in Art. 28. If
viewed so that the rotation of the angular impulse or momentum
228 APPLIED MECHANICS [CHAP, xn
appears negative (clockwise), the vector points away from the
observer. The vectors for Figs. 311 and 312 would be perpen-
dicular to the plane of the figures, and the arrow would point
toward the observer, since the rotation is counterclockwise. Like
other vector quantities, angular impulse and angular momentum
may be resolved into components or combined into resultants as
desired. ,
EXAMPLE.
A flywheel weighs 3200 pounds and is rotating at 125 r.p.m. Its radius of
gyration is 3.3 feet and its shaft is 6 inches in diameter. If the coefficient of
friction / at the bearing is 0.02, how many revolutions will it make before
coming to rest?
Solution: — Friction = 3200 X 0.02 = 64 Ibs.
The impulse of the friction = 64 t.
The angular impulse = 64JXi = 16*.
Angular impulse = change in angular momentum.
16 t = 7co.
16/ - W Aft* 32°° 125
~ kw
16 t = 14,180.
t = 885 sec. = 14.75 min.
The average speed of the flywheel is 62.5 r.p.m., so in 14.75 minutes it will
turn through 62.5 X 14.75 = 922 revolutions.
Problem 1. When another lubricant was used, the wheel referred to in the
Example above came to rest in 22 minutes, 13 seconds. What was the coeffi-
cient of friction? Ans. 0.0133.
Problem 2. A cast iron cylinder 1 foot in diameter and 4 inches long is
supported in bearings by means of a shaft 2 inches in diameter. The coefficient
of friction is 0.01. A force of 5.44 pounds vertically downward is applied to
a cord wrapped around the cylinder. What is the rim velocity after 10
seconds if it starts from rest? Ans. 28.6 ft. per sec.
Problem 3. If in Problem 2 the force is released at the end of 10 seconds,
how long will the cylinder rotate until the friction of the bearing brings it to
rest? Ans. 4 min. 26 sec.
148. Conservation of Angular Momentum. If during any
time t there is no external angular impulse on a body or system
of bodies with respect to any given axis, the angular momentum
with respect to that axis remains constant, irrespective of mutual
actions and reactions between the bodies or parts of bodies. Since
the internal forces always occur in pairs of equal and opposite forces
during the same time t, the impulse of each pair, / F dt — I F dt,
ART. 148] IMPULSE, MOMENTUM AND IMPACT 229
reduces to zero. Since the angular impulse is zero, there can be
no change in angular momentum.
Consider as an example two disks, A and B, Fig. 313, supported
on a horizontal shaft. Let disk A be fastened to the shaft and be
at rest, while disk B rotates upon the shaft with angular velocity co.
Let the moment of inertia of. disk B be /, and the moment of in-
ertia of the entire system be /'. Since
disk A is at rest, the angular mo-
mentum of the system is equal to the
angular momentum of B} or /co. If
now by some means the two disks are
fastened together, as by allowing a bolt
in B to drop into a hole in A, the two
will rotate together with a new angular velocity co'. The angular
momentum of the system is now /'co', and since it has not been
changed by the internal action and reaction of the bolt and the
disks,
/'co' = /co, so co' •= 77 co.
FIG. 313
Both /co and /'co' are represented by the same vector as shown
at the left of the figure.
As in the case of linear momentum, there is a loss of kinetic
energy due to the mutual action between the two disks.
As another case, consider the two bodies A and B, Fig. 314,
which may be moved along a horizontal axis
and which are rotating with their support
about a vertical axis with angular velocity co.
If / is their moment of inertia with respect to
their axis of rotation, their angular momen-
tum is /co. If now by some internal action
FIG. 314
the bodies are displaced, as to A' and B', the moment of inertia
of the system becomes I'. Then, as above, co' = -^ co. Since /' is
less than /, co' must be correspondingly greater than co.
Problem 1. In Fig. 313, the disks are steel, 18 inches in diameter. B is
2 inches thick and A is \ inch thick. If B is rotating originally at a speed of
60 r.p.m., what will be the speed of rotation after being connected? What is
the loss in kinetic energy by the blow as they are connected?
Ans. 53.4 r.p.m. 2.77 ft.-lbs.
230 APPLIED MECHANICS [CHAP.XU
149. Resultant of Angular Momenta. Gyroscope. A gyro-
scope is a body which is rotating about an axis, called the spin
axis, and is partially free to move in other directions. If the
armature of a motor, Fig. 315, is suspended from a support and is
rotating at a high rate of speed, it offers the same resistance to
any translatory force or to any torque about the spin axis that it
would if not rotating. If, however, a torque is applied to rotate
it about any other axis, as the vertical, it will not rotate about the
vertical axis, but will rotate or precess about a horizontal axis
perpendicular to the other two axes. In Fig. 315, OX is the spin
axis, OF is the torque axis and OZ, mutually perpendicular to
these, is the precession axis.
FIG. 315 FIG. 316
The explanation of this precession is as follows. If I is the
moment of inertia of the rotating part with respect to the axis
of rotation and co is its angular velocity, its angular momentum is
7co, represented by vector ON, Fig. 316. The vector points to
the left if the rotation is clockwise when viewed from the right
end. If a torque M is applied to the frame which tends to rotate
the motor about the axis OF in a clockwise direction viewed from
above, this torque will have an angular impulse Mdt in dt time,
and will produce an equal change in the angular momentum about
axis OF. This is represented by vector OL. The resultant
angular momentum of the body is. represented in amount and
direction by the resultant OP of the two angular momentum
vectors, ON and OL, at an angle d<j> with the vector ON. This is
the new axis of spin and in order that it shall become so the body
must rotate about axis OZ, the direction being counterclockwise
viewed from the front. This motion is called precession, and «'
is the angular velocity of precession.
tan d*t> = dtj>.
ART. 149] IMPULSE, MOMENTUM AND IMPACT 231
OL Mdt
ON~ /co '
Mdt
(MJ) = — — — «
/CO
dd) JKZ
Tt = J^'
But — = co', the angular velocity of precession about axis OZ, so
W = Ico'
M =Icoco'.
If the torque about axis OY remains constant, the velocity co'
about axis OZ remains constant. Since there is no acceleration
about axis OZ, Mz must equal zero.
Just as a torque about an axis normal to the spin axis causes a
precession about a third axis normal to the two, a forced pre-
cession about an axis normal to the spin axis will cause a torque
about the third rectangular axis. For example, if a uniform pre-
cession be caused about the axis OZ in a counterclockwise direction
viewed from the front, a torque M about the axis OF is developed.
This torque is in the clockwise direction, viewed from above, and
is given by the axle reactions.
EXAMPLE.
The armature of the motor of an electric car weighs 600 pounds and rotates
in the direction opposite to the rotation of the car wheels. The distance
between bearings is 2 feet and the radius of
gyration of the armature is 6 inches. The // 0 \w x lev
motor is geared so that it makes four revolu-
tions to one revolution of the car wheels. The
diameter of the car wheels is 33 inches. If the
car is going forward around a curve of 100 feet
radius with a velocity of 20 feet per second,
what are the pressures on the bearings if the
center of the curve is to the right ? fiW Rear V
Solution: — Fig. 317(a) is a top view and
Fig. 317(b) is a rear view of the motor. OX
is the spin axis, a vertical axis through the center of curvature of the
track is the precession axis and any axis normal to these is the torque axis.
Since the armature is being accelerated toward the center of the curve
232 APPLIED MECHANICS [CHAP, xn
4)2
with an acceleration a = — = 4 feet per sec. per sec., the horizontal pressure
of the bearing,
H = 95 x 4 = 74.5 ibs.
This is due to the centrifugal force and would be the same if the motor were
not rotating. Since there is no torque about the axis of precession, there are
no horizontal components of the reactions at the ends of the armature normal
to#.
Since the spin is backward, the angular momentum vector /co points to
the right. In order to combine with this momentum vector so as to produce
precession to the right, the angular impulse vector must point backward along
the track toward the observer. The torque to give the vector this direction
is counterclockwise, so Pi must be larger than PI.
Since M = /coco', by moments about 0,
P2 X 2 - 600 X 1 = /coco'.
The angular velocity of rotation of the car wheels is
coi = - = = 14.55 rad. per sec.
T l.oYO
Since the gear ratio is 4 to 1, the angular velocity co of the armature is
4 X 14.55 = 58.2 rad. per sec.
co' = - = -^ = 0.2 rad. per sec.
2 P2 - 600 = 4.66 X 58.2 X 0.2.
2 P2 - 600 = 54.
P2 = 327 Ibs.
Since PI + P2 = 600, P! = 273 Ibs.
It is seen that the pressure on the bearing on the inside of the curve is 27
Ibs. heavier than if the motor were not rotating and the pressure on the out-
side is 27 Ibs. lighter.
For a curve in the opposite direction the angular momentum vector would
still point to the right, and the angular impulse vector would have to point
forward in order that the two would combine to produce precession to the
left. The torque is therefore of opposite sign, so the heavier pressure is again
on the inside bearing.
For a motor geared so that it rotates in the same direction as the car wheels,
the heavier pressure is on the bearing on the outside of the curve.
Upon the wheels themselves there is a like gyroscopic action, so that the
pressure on the outer rail is greater and that on the inner rail is less than it
would be with no rotation.
Problem 1. In Problem 1, Art. 79, it was found that a pair of 33-inch cast
iron car wheels weighing 700 pounds had a moment of inertia of 6.99 with
respect to the axis of rotation. If a car is running at a speed of 30 miles per
ART. 151] IMPULSE, MOMENTUM AND IMPACT 233
hour around a 10° curve, what is the extra pressure on the outer rail due to
gyroscopic action? Use 4.9 feet as the distance from center to center of rail.
Ans. 3.54 Ibs. for each pair of wheels.
Problem 2. Describe the gyroscopic action of the flywheel of an automobile
engine when rounding a curve (1) to the right; (2) to the left.
Ans. (1) Heavier pressure on rear bearing.
Problem 3. If an ordinary top is rotating clockwise viewed from above
and the upper end of the axis is pushed horizontally north, which way will it
really lean? Explain. Ans. East.
Problem 4. If an aeroplane which is coming down head on at a steep angle
changes direction by a short curve into the horizontal, what will be the gyro-
scopic action if the propeller is rotating clockwise when viewed from the rear?
Ans. Left end will be thrown forward.
150. Reaction of a Jet of Water. If a jet of water of cross-
sectional area A issues from the side of a vessel of water under
head h, as shown in Fig. 318, it will have
a velocity v = ^/2gh. The water before
leaving the vessel is at rest, so the change
in momentum in the direction of the jet is
Mv per second, M being the mass of water
W
flowing per second. Mv = — .v, W being the FIG. 318
y
weight of water flowing per second. If w is the weight of the
unit volume of water, W = wAv. Then the change in momentum
wAv^
per second is Mv = - — . The change in momentum in time t
i/
7/) A ifi
is - -t and this must be caused by an impulse Ft. Then since
2
F = = 2 wAh.
g
F' is the equal and opposite reaction of the water upon the
vessel. If the vessel is not held by some external force, it will
move to the right under the action of force F' .
Problem 1. A cubical vessel 1 foot on each side, weighing 7.5 pounds is full
of water and is suspended from a cord so that its center of gravity is 5 feet
from the support. If a jet 1 square inch in area under a head of 1 foot is
issuing from the side, how far from the vertical is the center of gravity dis-
placed? Ans. 0.745 in.
151. Pressure Due to a Jet of Water on a Vane. If a jet of
water is discharged perpendicularly against a stationary flat vane,
234
APPLIED MECHANICS
[CHAP, xn
as in Fig. 319, all of the velocity of the jet in the original direction
is destroyed. If W is the weight of water flowing per second, the
W
change in momentum in the direction of the jet is — v per second.
W
The change in momentum in time t is — vt and this must equal
y
the impulse of the force F which supports the vane. Then
9
Wv
wAv2
- 0 AU
F = — = - = 2 wAh.
g g
It will be noticed that the pressure P due to an impinging jet
is twice as great as the static pressure P' = wAh on the same
area as the area of the jet, due to a head of water corresponding
to the same velocity.
D
FIG. 320
If a jet of water strikes a curved vane tangentially, as shown in
Fig. 320, and is turned through an angle 0, it will leave in a direc-
tion tangent to the vane at B with the same speed as that with
which it enters at A neglecting friction. The vector OC may be
drawn to represent to scale the velocity at A and OD to represent
the velocity at B. Then CD is the vector change in the velocity
of the jet. If W is the weight of water flowing per second, the
vector change in momentum in time t is
W — W fi
— CDt = —2vsm?t,
9 9 2
in the direction CD. This must equal Ft, since
Impulse = change in momentum.
The resultant pressure on the vane necessary to hold it against
the jet is, therefore,
™ w e
F = — 2 vsm--
g 2
ART. 152] IMPULSE, MOMENTUM AND IMPACT 235
Problem 1. What pressure will be exerted upon a flat vane held normal to
a jet of water 2 inches in diameter under a head of 60 feet? Ans. 164 Ibs.
Problem 2. A fire stream from a nozzle 1.5 inches in diameter has a velocity
of 120 feet per second. What pressure does it exert upon a wall at close
range? Ans. 343 Ibs.
Problem 3. What is the pressure in amount and direction exerted by a jet
of water 2 inches in diameter under a head of 120 feet upon a curved vane for
which e = 90°? Ans. 463 Ibs., 45° with initial v.
Problem 4. Solve Problem 3 if 0 = 180°. Ans. 655 Ibs. parallel to jet.
152. Sudden Impulse or Impact. The impulse of a force
which acts for a very short time is called a sudden impulse or
impact, as for example the blow of a hammer upon a nail, the col-
lision of a projectile and its target, the impact of a bat upon a
ball. If the mass centers of the two bodies before collision move
along the same straight line and the form of the bodies is such
that the pressure each exerts upon the other is also along this
line,^the impact is called direct central impact. All
other impacts are oblique.
For simplicity assume the colliding bodies to be
spheres, as in Fig. 321. The mass MI moving with
velocity v\ overtakes mass M2 moving with velocity
v2. When the bodies first touch, as in Fig. 321 (a),
the pressure between them is zero. For a short
period of time, the centers approach each other and
each is deformed by the pressure of the other.
When the pressure becomes a maximum the def-
ormation is a maximum, as shown in Fig. 321(b),
and the bodies are moving with the same velocity v. F
If the bodies are inelastic, the pressure drops di-
rectly to zero, the deformation remains and the two bodies go on
together with velocity v. By the principle of conservation of linear
momentum, Art. 146,
+ M2V2 = MIV + M2v.
If the bodies are partially elastic, the pressure decreases gradu-
ally to zero, the original form is partially regained and the two
bodies separate, MI moving with velocity v\ and M2 with velocity
Vz, as shown in Fig. 321 (c). In this case also, by Art. 146,
M,v + M2v = M,VI' + M2v2'.
The first period of time is called the period of compression. The
second period is called the period of restitution.
236
APPLIED MECHANICS
[CHAP, xn
From the two equations above the sum of the momenta before
impact equals the sum of the momenta after impact.
+ M2v2 = Miv/ + M2v2'.
The direction of v\ should always be considered positive. If v2
is in the opposite direction, it must be used as negative. The
signs of vi and v2 show their directions.
The relative velocity of the two bodies before impact is vi — v2
and the relative velocity after impact is v2 — vi. Due to the
fact that physical bodies are not perfectly elastic, the relative
velocity after impact is always less than that before impact, and
the ratio of the two is called the coefficient of restitution, represented
by e.
e =
— V2
This may be written
e (vi - v2) = v2' - V/.
The value of e is zero for entirely inelastic bodies and would be
unity for perfectly elastic bodies. The following table gives the
values of e for several materials as determined by experiment.
Material
e
Glass
0 95
Ivory . .
0 89
Steel
0.55
Cast iron
Lead ....
0.50
0.15
The kinetic energy lost in heat of impact may be found by
subtracting the final kinetic energy of the two bodies from their
initial kinetic energy.
If the impact of two bodies is oblique, the velocity of each body
may be resolved into two components, one along the line of
centers, the other normal to the line of centers. The latter com-
ponent of each is unchanged by the impact. The former is
changed the same as in direct central impact and the final com-
ponents are recombined to give the final velocities.
EXAMPLE 1.
An inelastic body weighing 5 pounds is moving with a velocity of 10 feet
per second and collides with another weighing 2 pounds moving in the opposite
direction with a velocity of 6 feet per second. What is the final velocity of
the two bodies and the loss in kinetic energy due to the impact?
ART. 152] IMPULSE, MOMENTUM AND IMPACT 237
Solution: — M\VI f M2v2 = Mtv + M2v.
5 2 /5 2\
9 9 ~ \9 9JV'
v = 5.43 ft. per sec.
Initial K.E. = ^ ^ X 100 + i ^ X 36 = 8.88 ft.-lbs.
Final K.E. - 1 o^ X 5.432 = 3.21 ft.-lbs.
Loss in K.E. = 8.88- 3.21 = 5.67 ft.-lbs.
EXAMPLE 2.
A 10-pound steel ball at rest is struck horizontally by a hammer weighing
4 pounds with a velocity of 12 feet per second. What is the velocity of each
after the impact and the loss in kinetic energy?
Solution: — The hammer is Mi and the ball M2. From the table above,
e = 0.55.
0.55 (12— 0) = v2'— Vi' = 6.6.
48 + 0 = 4 vi' + 10 v2'.
vi' = — 1.29 ft. per sec.
v2' = 5.31 ft. per sec.
The ball is driven in the direction the hammer was going originally; the
hammer itself rebounds.
Initial K.E. = i ^ 144 = 8'95 ft'"lbs'
Final K.E. of hammer = | ^ L298 = 0'10 ft'"lb'
Final K.E. of ball = i -^ 5.312 = 4.39 ft.-lbs.
Loss in K.E. = 8.95- 4.49 = 4.46 ft.-lbs.
Problem 1. An inelastic body weighing 10 pounds is moving with a
velocity of 20 feet per second and overtakes another of 15 pounds weight
moving in the same direction with a velocity of 15 feet per second. What is
the final velocity and the loss in kinetic energy?
Ans. 17 ft. per sec. 2.4 ft.-lbs.
Problsir 2. An inelastic ball of 30 pounds weight moving with a velocity
of 100 feet per second strikes another of 30 pounds weight which is at rest.
What is the final velocity and the loss in kinetic energy?
Ans. 50 ft. per sec. 2330 ft.-lbs.
Problem 3. A glass marble weighing 2 ounces drops 2 inches upon a heavy
glass slab. To what height will it rebound? What is the loss in kinetic
energy due to the impact? Ans. 1.8 in. 0.002 ft.-lb.
Problem 4. A cast iron ball weighing 1 pound is moving with a speed of
10 feet per second and strikes another weighing 7.5 Ibs. moving in the oppo-
site direction with a speed of 3 feet per second. What are the final velocities?
Ans. vi = —7.21 ft. per sec. v2 = —0.71 ft. per sec.
Problem 5. With what velocity must a 5-pound steel hammer strike a 1-
pound steel ball at rest, in order to drive it with a velocity of 100 feet per
second? Ans. 77.5 ft. per sec.
238 APPLIED MECHANICS [CHAP, xn
GENERAL PROBLEMS.
Problem 1. A freight car weighing 40,000 pounds starts from rest and runs
down a 2 per cent grade. If train resistance is 6 pounds per ton, what is its
velocity at the end of one minute? Am. 32.8 ft. per sec.
Problem 2. If the train resistance on the car described in Problem 1 is
6 + 0.1 t pounds per ton, t being in seconds, what is its velocity at the end of
one minute? Ans. 30 ft. per sec.
Problem 3. If at the end of one minute brakes are applied so as to stop the
car described in Problem 1 in 10 seconds, what is the braking force required?
Ans. 4760 Ibs.
Problem 4. A gun weighing 5 pounds shoots a bullet weighing 0.1 ounce
with a muzzle velocity of 800 feet per second. What is the recoil velocity of
the gun? Ans. 1 ft. per sec.
Problem 5. A push car weighing 300 pounds is moving with a uniform
velocity of 15 feet per second. If a man weighing 150 pounds boards it from
the side with no velocity in the direction it is moving, what is their velocity
after the man comes to rest with respect to the car?
Ans. v = 10 ft. per sec.
Problem 6. A mine cage weighs 800 pounds and the drum of the hoisting
engine connected to it weighs 1000 pounds. If A; is 1.5 feet, the diameter of
the drum is 4 feet, the diameter of the shaft is 2 inches and the coefficient of
friction at the bearings is 0.03, what velocity will the cage attain if allowed to
fall for 5 seconds before the brake is applied? Ans. 94.3 ft. per sec.
Problem 7. Two cast iron spheres 4 inches in diameter connected by a
slender rod with their centers 3 feet apart are rotating in a horizontal plane
about an axis normal to the rod at its middle at a speed of 60 r.p.m. If by
means of a spring joining them the two spheres are brought into contact at the
middle, what will be their new speed of rotation? Ans. 3480 r.p.m.
Problem 8. If a spherical top 2 inches in diameter weighs 1 pound and is
spinning at a speed of 800 r.p.m., what will be its time of precession if its point
is 1.2 inches from the center of gravity and its axis is 15° from the vertical?
Ans. 1.75 sec.
Problem 9. A gyroscope is composed of a circular rim weighing 10 pounds
on a bicycle wheel 24 inches in diameter. The wheel is on the end of a hori-
zontal axis 3 feet long, supported at its middle on a pivot so that it is free to
rotate in any direction. If the wheel is rotating about its own axis with a
speed of 600 r.p.m. clockwise when viewed from the pivot, in which direction
will it precess and with what angular velocity?
Ans. Counterclockwise, viewed from above, co' = 0.769 rad. per sec.
Problem 10. Discuss the gyroscopic action of the propeller of a ship as the
ship pitches fore and aft on the waves.
Problem 11. If the coefficient ol restitution e = 0.9 for a rubber ball, how
high will it rebound if dropped from a height of 10 feet? Ans. 8.1 ft.
Problem 12. A cast iron sphere 2 inches in diameter when dropped from
a height of 12 inches upon a cast iron block rebounded 4 inches. Compute e
and the loss in kinetic energy. Ans. e = 0.577. 0.727 ft.-lb.
IMPULSE, MOMENTUM AND IMPACT 239
Problem 13. A 100,000-pound railroad car moving with a speed of 5 miles
per hour overtakes and collides with another weighing 90,000 pounds moving
with a speed of 2 miles per hour. What is the loss in kinetic energy if e = 0.20?
Ans. 13,080 ft.-lbs.
Problem 14. From a point 5 feet above the ground a ball is thrown upward
at an angle of 60° with the horizontal against a wall 20 feet distant: If its
initial velocity is 60 feet per second and e for the ball is 0.50, where and with
what velocity will the ball strike the ground?
Ans. 39.8 ft. from wall, v = 86.7 ft. per sec., 10° with vertical.
Problem 15. If the ball described in Problem 14 is thrown horizontally
with the same velocity, but at an angle of 60° with the wall, where will it strike
the wall? Where will it strike the ground and with what velocity?
Ans. 10 ft. along the wall, 2.61 ft. from ground. 15.13 ft. along the wall,
4.45 ft. from the wall, v = 43.5 ft. per sec. Angle with ground = 24° 20'.
Angle with wall = 41°.
INDEX
Absorption dynamometer, 217.
Acceleration, 120.
angular, 155, 159, 160.
constant, 121.
in curvilinear motion, 140.
normal component of, 141.
tangential component of, 141.
variable, 132, 133.
Accelerations, composition of, 128.
Algebraic method of analysis, 2.
Amplitude, 145, 160.
Angle, of friction, 83.
of repose, 83.
Angular, acceleration, 155, 159, 160.
displacement, 154.
impulse, 226.
momentum, 226.
velocity, 154.
Area, centroid of, 66, 69.
moment of inertia of, 99.
Auxiliary circle, 156.
Axes of symmetry, 67.
Axis, inertia, 99.
instantaneous, 182.
Axle friction, 86.
Balancing, 173, 174, 175.
of locomotives, 195.
reciprocating parts, 193.
both rotating and reciprocating
parts, 195.
Band brakes, 219.
Belt friction, 94.
Brake friction, 90.
Braking of trains, 213.
Brakes, band, 219.
Built-up sections, moment of inertia
of, 116.
Catenary, 56.
Center, of oscillation, 169.
of percussion, 171.
Center of gravity, 65.
of composite body, 75.
by experiment, 77.
Centrifugal, force, 143.
tension in flywheels, 171.
Centroid, of a force system, 65, 68.
of irregular plane area, 77.
of a line, 66.
of a solid, 66.
of a surface, 66.
Centroids, 65.
by integration, 70.
of surfaces and solids of revolu-
tion, 73.
Circle, friction, 86.
Circular pendulum, 145.
Coefficient, of kinetic friction, 83,
85.
of restitution, 236.
of rolling resistance, 90.
of static friction, 83, 85.
Combined translation and rotation,
179.
Composition, of accelerations, 128.
of couples, 31.
of forces, 39, 40, 49.
of velocities, 128.
Compound pendulum, 169.
Compression, period of, 235.
Concurrent system of forces, 5.
Cone of friction, 87.
Conical pendulum governor, 172.
Connecting rod of engine, 187.
Conservation, of angular momentum,
228.
of linear momentum, 225.
241
242
INDEX
Cord loaded uniformly horizontally,
53.
Couple, arm of, 29.
graphic representation of, 30.
moment of, 29.
Couples, composition of, 31.
Curvilinear motion, 140.
D'Alembert's principle, 125.
Direct central impact, 235.
Dynamometer, absorption, 217.
Effective forces, 125.
on a rotating body, 143, 160.
moment of tangential, 161.
resultant of normal, 164.
Efficiency, mechanical, 215, 219.
Energy, 202.
kinetic, 203.
Equations of motion, 183.
Equilibrium, of colinear forces, 6.
of concurrent forces in space, 13,
16.
of coplanar forces, 8, 11.
of parallel forces, 32, 34.
of parallel forces in space, 35.
problems in, 44.
Falling bodies, 122.
Flywheels, centrifugal tension in, 171.
Force, centrifugal, 143.
concentrated, 3.
definition of, 1.
diagram, 4.
distributed, 3.
effect of, 124.
moment of, 17, 18.
polygon, 8, 24.
projection of, 7.
resolution into another force and a
couple, 32.
unit of, 1.
Forces, composition of, 39, 49.
concurrent system of, 5.
effective, 125.
effective, on a rotating body, 160.
equilibrium of, 40, 41.
graphical representation of, 4.
Forces, moment of tangential effec-
tive, 161.
parallel, 23.
parallel in equilibrium, 32, 34, 35.
reduction of a system to a force and
a couple, 41.
resolution and recomposition of,
11.
resultant of normal effective, 164.
transmissibility of, 4.
Free-body diagram, 3.
Friction, 82.
angle of, 83.
axle, 86.
belt, 94.
brake, 90.
circle, 86.
coefficient of, 83, 85.
cone of, 87.
kinetic, 82.
laws of, 85.
limiting, 82.
pivot, 92.
rolling, 90.
static, 82.
summary of principles of, 96.
Funicular polygon, 24.
Governor, weighted conical pendu-
lum, 172.
Graphic method of analysis, 2.
Graphical representation, of accelera-
tion, 121.
of angular impulse, 227.
of angular momentum, 227.
of a couple, 30.
of a force, 4.
of impulse, 223.
of momentum, 223.
of velocity, 119.
of work, 201.
Gravity, center of, 65.
Gyration, radius of, 100
Gyroscope, 230.
Harmonic motion, simple, 129, 156.
Horsepower, 215.
INDEX
243
Impact, 235.
direct central, 235.
oblique, 235.
Impulse, 223, 224,
angular, 226.
Indicator, steam engine, 216.
Inertia, axis, 99.
moment of, 99.
product of, 107.
instantaneous axis, 182.
Jet of water, 233.
Kinematic discussion, 1.
Kinetic energy, of rotation, 208.
of rotation and translation, 210.
of translation, 203.
Kinetic friction, 82.
Kinetic reaction on unbalanced wheel,
192.
Kinetic reactions, on connecting rod,
189.
on side rod, 191.
Kinetics, 1, 119.
Laws, of friction, 85.
of motion, Newton's, 3.
Line, centroid of, 66.
moment of, with respect to a plane,
66.
Mass, definition, 1.
unit, 2.
Mechanical efficiency, 215, 219.
Mechanics, 1.
divisions of, 1.
fundamental quantities in, 1.
Methods of analysis of problems, 2.
Moment, of a couple, 29.
of a force with respect to a line, 18.
of a force with respect to a plane,
28.'
of a force with respect to a point,
17.
of a solid, surface or line with re-
spect to a plane, 66.
of tangential effective forces, 161.
sign of, 17.
Moment, unit of, 17.
Moment of inertia, of an area, 99.
of an area, polar, 100.
of an area, polar, with respect to
parallel axes, 104.
of built-up sections, 116.
of composite area, 105.
by experiment, 115.
of mass, 111.
maximum and minimum, 110.
with respect to inclined axes, 106.
with respect to parallel axes, 102.
with respect to three rectangular
axes, 104.
sign of, 102.
Moments, principle of, 17.
Momentum, 223, 224.
angular, 226.
Motion, curvilinear, 140.
equations of, 183.
Newton's laws of, 3.
rectilinear, 119.
relative, 134.
simple harmonic, 129, 156.
Newton's laws of motion, 3.
Normal, acceleration, 141.
effective forces, 164.
Oblique impact, 235.
Oscillation, center of, 169.
Pappus and Guldinus, theorems of,
74.
Parallel forces, 23.
in equilibrium, 32, 34, 35.
Parallelogram law, 5.
Pendulum, conical, 172.
compound, 169.
simple circular, 145.
Percussion, center of, 171.
Period, of compression, 235.
of restitution, 235.
Pivot friction, 92.
Planes of symmetry, 67.
Polar moment of inertia of an area,
100.
Polygon of forces, 8, 24.
244
INDEX
Potential energy, 203.
Power, 215.
Pressure of water on a vane, 233.
Principle of moments, 17, 25, 40.
Principles of friction, summary of, 96.
Problems in equilibrium, 44, 50.
Product of inertia, 107.
Projectile, 149.
Projection, of a force, 7.
of a force polygon, 13.
of a force system, 16.
Radius of gyration, of an area, 100.
of mass, 111.
Reactions of supports of rotating
bodies, 166.
Rectilinear motion, 119.
Reference planes, 13.
Relative motion, 134.
Resistance, rolling, 89.
Resolution, of accelerations, 128,
181.
of a force into a force and a couple,
32.
of a force into three components,
14.
of a force into two components, 7.
of velocities, 128, 180.
Restitution, period of, 235.
coefficient of, 236.
Resultant, of angular momenta, 230.
of normal effective forces, 164.
of three or more forces, 8, 10, 13, 15.
of three or more parallel forces, 26,
27.
of two concurrent forces, 5, 6.
of two parallel forces, 23, 24.
Rolling resistance, 89.
coefficient of, 90.
Rotation, 154.
kinetic energy of, 208.
and translation combined, 179.
Simple circular pendulum, 145.
Simple harmonic motion, 129, 156.
Solid, centroid of, 66, 69.
moment of, with respect to a plane,
66.
Space, diagram, 4.
units of, 1.
Speed, 119.
Static friction, 82.
Statics, 1.
Steam engine indicator, 216.
Surface, centroid of, 66, 69.
moment of, with respect to a plane,
66.
Symmetry, axes of, 67.
planes of, 67.
Tangential acceleration, 141.
Three-force members, 42.
Time, unit of, 1.
Translation and rotation combined,
179.
Triangle law, 5, 8.
Two-force members, 42.
Uniform, circular motion, 142.
velocity, 119.
Variable, acceleration, 120.
velocity, 119.
Varignon's theorem, 18.
Vectors, 2.
Velocities, composition of, 128.
Velocity, 119.
angular, 154.
uniform, 119.
variable, 119.
in a vertical curve, 147.
Water, jets, 233. N
power, 216.
Weighted conical pendulum governor,
172.
Work, 199.
graphical representation of, 201.
lost in friction, 211.
UNIVERSITY OF CALIFORNIA LIBRARY
This book is DUE on the last date stamped below.
k 25 cents on first clay overdue
day overdue
cfav overdue.
UCT 6 1947
DEC 8 1947
MAY jo
** *S48
JUN 4 1348
31Jan'49Hl
REC'D
OCT271959
LD 21-100m-12,'46(A2012sl6)4120
THE UNIVERSITY OF CALIFORNIA LIBRARY
Live Music
Archive
Librivox
Free Audio
Metropolitan Museum
Cleveland
Museum of Art
Internet
Arcade
Console Living Room
Open Library
American
Libraries
TV News
Understanding
9/11